Antiderivatives

Exercises · Integrals · with worked solutions · PDF

Exercise 1

Compute the following antiderivatives on the intervals where they are defined

\[ {\rm (a)}\ \int\left(2x+\sqrt[3]{x}-\frac{2}{\sqrt{x}}\right)dx, ~~~{\rm (b)}\ \int\frac{x-x^{4}}{\sqrt{x}}dx \]

\[ {\rm (c)}\ \int\tan^{2}x\ dx,~~~~ {\rm (d)}\ \int(3-x)^{5}dx \]

\[ {\rm (e)}\ \int\frac{1}{2-3x}dx, ~~~{\rm (f)}\ \int x^{2}e^{x^{3}+1}dx \]

\[ {\rm (g)}\ \int x\sqrt{x^{2}+1}dx, ~~~~{\rm (h)}\ \int\frac{\sin2x}{1+\sin^{2}x}dx \]

\[ {\rm (i)}\ \int\frac{e^{x}}{e^{x}+3}dx, ~~~{\rm (j)}\ \int\frac{1}{\tan x} dx \]

Solution

(a)

\[ \begin{array}{l}\ds\int\left(2x+\sqrt[3]{x}-\frac{2}{\sqrt{x}}\right)dx=\int(2x+x^{1/3}-2x^{-1/2})dx\\ \\ \ds=x^{2}+\frac{3}{4}x^{4/3}-4x^{1/2}+c=x^{2}+\frac{3}{4}\sqrt[3]{x^{4}}-4\sqrt{x}+c\end{array} \]

on any interval \(I\) such that \(I\subset(0,+\infty)\).

Solution

(b)

\[ \begin{array}{l}\ds\int\frac{x-x^{4}}{\sqrt{x}}dx=\int(x^{1/2}-x^{7/2})dx=\\ \\ \ds\frac{2}{3}x^{3/2}-\frac{2}{9}x^{9/2}+c=\frac{2}{3}x\sqrt{x}-\frac{2}{9}x^{4}\sqrt{x}+c\end{array} \]

on any interval \(I\) such that \(I\subset(0,+\infty)\).

Solution

(c)

\[ \int\tan^{2}x\ dx=\int[(1+\tan^{2}x)-1]dx=\tan x-x+c \]

on any interval \(I\) that does not contain points of the form \(\pi/2+k\pi\) with \(k\in \Z\).

Solution

(d)

\[ \int(3-x)^{5}dx=-\int(x-3)^{5}dx=-\frac{1}{6}(x-3)^{6}+c \]

on any interval \(I\) of \(\R\).

Solution

(e)

\[ \int\frac{1}{2-3x}dx=-\int\frac{1}{3x-2}dx=-\frac{1}{3}\log|3x-2|+c \]

on any interval \(I\) such that \(2/3\notin I\).

Solution

(f)

\[ \int x^{2}e^{x^{3}+1}dx=\frac{1}{3}e^{x^{3}+1}+c \]

on any interval \(I\) of \(\R\).

Solution

(g)

\[ \int x\sqrt{x^{2}+1}dx=\frac{1}{3}(x^{2}+1)^{3/2}+c=\frac{1}{3}(x^{2}+1)\sqrt{x^{2}+1}+c \]

on any interval \(I\) of \(\R\).

Solution

(h)

\[ \int\frac{\sin2x}{1+\sin^{2}x}dx=\int\frac{2\sin x\cos x}{1+\sin^{2}x}dx=\log(1+\sin^{2}x)+c \]

on any interval \(I\) of \(\R\).

Solution

(i)

\[ \int\frac{e^{x}}{e^{x}+3}dx=\log(e^{x}+3)+c \]

on any interval \(I\) of \(\R\).

Solution

(j)

\[ \int\frac{1}{\tan x} dx=\int\frac{\cos x}{\sin x}dx=\log|\sin x|+c \]

on any interval \(I\) that does not contain points of the form \(k\pi\) with \(k\in \Z\).

Exercise 2

Compute the following antiderivatives using integration by parts

\[ {\rm (a)}\ \int x\sin x\ dx, ~~~~{\rm (b)}\ \int x^{2}\cos x\ dx \]

\[ {\rm (c)}\ \int\arcsin x\ dx, ~~~~{\rm (d)}\ \int\sin^{2}x\ dx \]

Solution

(a)

\[ \int x\sin x\ dx=-x\cos x+\int \cos x\ dx=-x\cos x+\sin x+c \]

on any interval \(I\) of \(\R\).

Solution

(b) Integrating by parts twice in a row, we have

\[ \begin{array}{l}\ds\int x^{2}\cos x\ dx=x^{2}\sin x-2\int x\sin x\ dx=\\ \\ \ds x^{2}\sin x+2x\cos x-2\int\cos x\ dx=x^{2}\sin x+2x\cos x-2\sin x+c\end{array} \]

on any interval \(I\) of \(\R\).

Solution

(c) Taking the constant function \(1\) as the factor to be integrated, we have

\[ \int\arcsin x\ dx=x\arcsin x-\int\frac{x}{\sqrt{1-x^{2}}}dx=x\arcsin x+\sqrt{1-x^{2}}+c \]

on any interval \(I\subset(-1,1)\).

Solution

(d) Integrating by parts and using \(\cos^{2}x=1-\sin^{2}x\), we obtain

\[ \begin{array}{l}\ds\int\sin^{2}x\ dx=-\sin x\cos x+\int\cos^{2}x\ dx\\ \\ \ds=-\sin x\cos x+\int 1\ dx-\int\sin^{2}x\ dx\end{array} \]

from which

\[ \int\sin^{2}x\ dx=-\sin x\cos x+x-\int\sin^{2}x\ dx+c \]

hence, moving \(\int\sin^{2}x\ dx\) to the left-hand side,

\[ 2\int\sin^{2}x\ dx=-\sin x\cos x+x+c \]

and finally, still denoting by \(c\) the arbitrary constant \(c/2\),

\[ \int\sin^{2}x\ dx=-\frac{1}{2}\sin x\cos x+\frac{1}{2}x+c \]

on any interval \(I\) of \(\R\).

Exercise 3

Compute the following antiderivatives using integration by substitution

\[ {\rm (a)}\ \int\frac{e^{x}-1}{e^{x}+1}dx,~~~ {\rm (b)}\ \int\frac{\sqrt{x}}{1+x}dx,~~~{\rm (c)}\ \int x\sqrt{x+1}dx \]

Solution

(a) Setting \(y=e^{x}\), \(y>0, x\in\R\), we have \(x=\log y\) and we write \(dx=\frac{1}{y}dy\), hence

\[ \int\frac{e^{x}-1}{e^{x}+1}dx=\int\frac{y-1}{y(y+1)}dy. \]

Using partial fractions, we obtain

\[ \frac{y-1}{y(y+1)}=-\frac{1}{y}+\frac{2}{y+1} \]

from which, taking into account that \(y>0\),

\[ \int\frac{y-1}{y(y+1)}dy=-\log y+2\log(y+1)+c. \]

It follows, substituting back \(y=e^{x}\),

\[ \int\frac{e^{x}-1}{e^{x}+1}dx=-\log e^{x}+2\log(e^{x}+1)+c=-x+2\log(e^{x}+1)+c \]

on any interval \(I\) of \(\R\).

Solution

(b) Setting \(y=\sqrt{x}\), \(x,y>0\), we have \(x=y^{2}\) and we write \(dx=2y\ dy\), hence

\[ \int\frac{\sqrt{x}}{1+x}dx=2\int\frac{y^{2}}{1+y^{2}}dy. \]

From

\[ \frac{y^{2}}{y^{2}+1}=\frac{y^{2}+1-1}{y^{2}+1}=1-\frac{1}{y^{2}+1} \]

we then have

\[ 2\int\frac{y^{2}}{1+y^{2}}dy=2y-2\arctan y+c. \]

It follows, substituting back \(y=\sqrt{x}\),

\[ \int\frac{\sqrt{x}}{1+x}dx=2\sqrt{x}-2\arctan\sqrt{x}+c \]

on any interval \(I\subset(0,+\infty)\).

Solution

(c) Setting \(y=\sqrt{x+1}\), \(x>-1,y>0\), we have \(x=y^{2}-1\) and we write \(dx=2y\ dy\), hence

\[ \int x\sqrt{x+1}dx=2\int(y^{2}-1)y^{2}\ dy=2\int(y^{4}-y^{2})dy=\frac{2}{5}y^{5}-\frac{2}{3}y^{3}+c. \]

It follows, substituting back \(y=\sqrt{x+1}\),

\[ \int x\sqrt{x+1}dx=\frac{2}{5}(x+1)^{5/2}-\frac{2}{3}(x+1)^{3/2}+c \]

on any interval \(I\subset(-1,+\infty)\).

Exercise 4

Compute the following antiderivatives of rational functions

\[ {\rm (a)}\ \int\frac{5}{x^{2}+2x+1}dx,~~~ {\rm (b)}\ \int\frac{1}{x^{2}+x+1}dx,~~~ {\rm (c)}\ \int\frac{x}{x^{2}+x+1}dx,~~~ {\rm (d)}\ \int\frac{x^{3}+1}{x^{2}-3x+2}dx \]

\[ {\rm (e)}\ \int\frac{x}{x^{3}+3x^{2}+3x+1}dx,~~~ {\rm (f)}\ \int\frac{1}{x^{4}-1}dx,~~~ {\rm (g)}\ \int\frac{x+1}{x^{3}-x^{2}}dx,~~~ {\rm (h)}\ \int\frac{1}{x^{3}+1}dx \]

Solution

(a) The given integral is immediate:

\[ \int\frac{5}{x^{2}+2x+1}dx=\int\frac{5}{(x+1)^{2}}dx=-\frac{5}{x+1}+c \]

on any interval \(I\) such that \(-1\notin I\).

Solution

(b) Writing the (irreducible) denominator as a sum of squares:

\[ \int\frac{1}{x^{2}+x+1}dx=\int\frac{1}{(x+1/2)^{2}+3/4}dx= \frac{2}{\sqrt{3}}\arctan\left[\frac{2}{\sqrt{3}}\left(x+\frac{1}{2}\right)\right]+c \]

on any interval \(I\) of \(\R\).

Solution

(c) Using also the previous exercise:

\[ \begin{array}{l}\ds\int\frac{x}{x^{2}+x+1}dx=\frac{1}{2}\int\frac{2x+1}{x^{2}+x+1}dx-\frac{1}{2}\int\frac{1}{x^{2}+x+1}dx=\\ \\ \ds\frac{1}{2}\log(x^{2}+x+1)-\frac{1}{\sqrt{3}}\arctan\left[\frac{2}{\sqrt{3}}\left(x+\frac{1}{2}\right)\right]+c\end{array} \]

on any interval \(I\) of \(\R\).

Solution

(d) We perform polynomial long division:

\[ \frac{x^{3}+1}{x^{2}-3x+2}=x+3+\frac{7x-5}{x^{2}-3x+2}. \]

Setting up the partial fractions

\[ \frac{7x-5}{x^{2}-3x+2}=\frac{7x-5}{(x-1)(x-2)}=\frac{A}{x-1}+\frac{B}{x-2}, \]

we obtain

\[ A=-2,\ B=9. \]

In conclusion:

\[ \begin{array}{l}\ds\int\frac{x^{3}+1}{x^{2}-3x+2}dx=\int\left(x+3-\frac{2}{x-1}+\frac{9}{x-2}\right)dx=\\ \\ \ds\frac{1}{2}x^{2}+3x-2\log|x-1|+9\log|x-2|+c\end{array} \]

on any interval \(I\) such that \(1\notin I\), \(2\notin I\).

Solution

(e) We set up the partial fractions:

\[ \frac{x}{x^{3}+3x^{2}+3x+1}=\frac{x}{(x+1)^{3}}=\frac{A}{x+1}+\frac{B}{(x+1)^{2}}+\frac{C}{(x+1)^{3}}. \]

We obtain

\[ A=0,\ B=1,\ C=-1. \]

In conclusion

\[ \begin{array}{l}\ds\int\frac{x}{x^{3}+3x^{2}+3x+1}dx=\int\left(\frac{1}{(x+1)^{2}}-\frac{1}{(x+1)^{3}}\right)dx\\ \\ \ds=-\frac{1}{x+1}+\frac{1}{2}\frac{1}{(x+1)^{2}}+c\end{array} \]

on any interval \(I\) such that \(-1\notin I\).

Solution

(f) We set up the partial fractions:

\[ \frac{1}{x^{4}-1}=\frac{1}{(x-1)(x+1)(x^{2}+1)}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{Cx+D}{x^{2}+1}. \]

We obtain

\[ A=\frac{1}{4},\ B=-\frac{1}{4},\ C=0,\ D=-\frac{1}{2}. \]

In conclusion

\[ \begin{array}{l}\ds\int\frac{1}{x^{4}-1}dx=\int\left(\frac{1}{4}\frac{1}{x-1}-\frac{1}{4}\frac{1}{x+1}-\frac{1}{2}\frac{1}{x^{2}+1}\right)dx\\ \\ \ds=\frac{1}{4}\log|x-1|-\frac{1}{4}\log|x+1|-\frac{1}{2}\arctan x+c\end{array} \]

on any interval \(I\) such that \(\pm1\notin I\).

Solution

(g) We set up the partial fractions:

\[ \frac{x+1}{x^{3}-x^{2}}=\frac{x+1}{x^{2}(x-1)}=\frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{x-1}. \]

We obtain

\[ A=-2,\ B=-1,\ C=2. \]

In conclusion

\[ \begin{array}{l}\ds\int\frac{x+1}{x^{3}-x^{2}}dx=\int\left(-\frac{2}{x}-\frac{1}{x^{2}}+\frac{2}{x-1}\right)dx\\ \\ \ds=-2\log|x|+\frac{1}{x}+2\log|x-1|+c\end{array} \]

on any interval \(I\) such that \(0,1\notin I\).

Solution

(h) We set up the partial fractions:

\[ \frac{1}{x^{3}+1}=\frac{1}{(x+1)(x^{2}-x+1)}=\frac{A}{x+1}+\frac{Bx+C}{x^{2}-x+1}. \]

We obtain

\[ A=\frac{1}{3},\ B=-\frac{1}{3},\ C=\frac{2}{3}. \]

Hence

\[ \begin{array}{l}\ds\int\frac{1}{x^{3}+1}dx=\int\left(\frac{1}{3}\frac{1}{x+1}+\frac{1}{3}\frac{-x+2}{x^{2}-x+1}\right)dx\\ \\ \ds=\frac{1}{3}\log|x+1|-\frac{1}{6}\int\frac{2x-1}{x^{2}-x+1}dx+\frac{1}{2}\int\frac{1}{x^{2}-x+1}dx\\ \\ \ds=\frac{1}{3}\log|x+1|-\frac{1}{6}\log(x^{2}-x+1)+\frac{1}{2}\int\frac{1}{(x-1/2)^{2}+3/4}dx\\ \\ \ds=\frac{1}{3}\log|x+1|-\frac{1}{6}\log(x^{2}-x+1)+\frac{1}{\sqrt{3}}\arctan\left[\frac{2}{\sqrt{3}}\left(x-\frac{1}{2}\right)\right]+c \end{array} \]

on any interval \(I\) such that \(-1\notin I\).

Exercise 5

Compute the following antiderivatives

\[ {\rm (a)}\ \int\frac{e^{x}+1}{e^{2x}+1}dx,~~~ {\rm (b)}\ \int x^{2}\log(x+1)dx,~~~ {\rm (c)}\ \int\frac{\tan x}{\sin^{2}x-\cos^{2}x}dx \]

\[ {\rm (d)}\ \int\frac{x+\sqrt{x}}{2+\sqrt{x}}dx,~~~ {\rm (e)}\ \int\frac{1}{\cos x}dx,~~~ {\rm (f)}\ \int\sqrt{4-x^{2}}dx \]

Solution

(a) With the substitution \(y=e^{x}\), \(y>0\), \(x\in\R\), we have \(x=\log y\), we write \(dx=\frac{1}{y}dy\), and we obtain

\[ \int\frac{e^{x}+1}{e^{2x}+1}dx=\int\frac{y+1}{y(y^{2}+1)}dy. \]

We set up the partial fractions:

\[ \frac{y+1}{y(y^{2}+1)}=\frac{A}{y}+\frac{By+C}{y^{2}+1}. \]

We have

\[ A=1,\ B=-1,\ C=1, \]

from which, taking into account also that \(y>0\),

\[ \begin{array}{l}\ds\int\frac{y+1}{y(y^{2}+1)}dy=\int\left(\frac{1}{y}-\frac{y}{y^{2}+1}+\frac{1}{y^{2}+1}\right)dy\\ \\ \ds=\log y-\frac{1}{2}\log(y^{2}+1)+\arctan y+c.\end{array} \]

Then, substituting back \(y=e^{x}\), we have

\[ \int\frac{e^{x}+1}{e^{2x}+1}dx=x-\frac{1}{2}\log(e^{2x}+1)+\arctan e^{x}+c \]

on any interval \(I\) of \(\R\).

Solution

(b) Integrating by parts and then computing the integral of a rational function, we have

\[ \begin{array}{l}\ds\int x^{2}\log(x+1)dx=\frac{1}{3}x^{3}\log(x+1)-\frac{1}{3}\int\frac{x^{3}}{x+1}dx\\ \\ \ds=\frac{1}{3}x^{3}\log(x+1)-\frac{1}{3}\int(x^{2}-x+1)dx+\frac{1}{3}\int\frac{1}{x+1}dx\\ \\ \ds=\frac{1}{3}x^{3}\log(x+1)-\frac{1}{9}x^{3}+\frac{1}{6}x^{2}-\frac{1}{3}x+\frac{1}{3}\log(x+1)+c \end{array} \]

on any interval \(I\subset(-1,+\infty)\).

Solution

(c) From \(\tan x=\frac{\sin x}{\cos x}\) and \(\sin^{2}x=1-\cos^{2}x\), we have

\[ \int\frac{\tan x}{\sin^{2}x-\cos^{2}x}dx=\int\frac{\sin x}{\cos x(1-2\cos^{2}x)}dx. \]

Now setting \(y=\cos x\), we write \(dy=-\sin x\ dx\) and we obtain

\[ \int\frac{\sin x}{\cos x(1-2\cos^{2}x)}dx=\frac{1}{2}\int\frac{1}{y(y-\sqrt{1/2})(y+\sqrt{1/2})}dy. \]

Using partial fractions, we then have

\[ \frac{1}{y(y-\sqrt{1/2})(y+\sqrt{1/2})}=-\frac{2}{y}+\frac{1}{y-\sqrt{1/2}}+\frac{1}{y+\sqrt{1/2}}, \]

from which

\[ \frac{1}{2}\int\frac{1}{y(y-\sqrt{1/2})(y+\sqrt{1/2})}=-\log|y|+\frac{1}{2}\log|y^{2}-1/2|+c. \]

In conclusion,

\[ \int\frac{\tan x}{\sin^{2}x-\cos^{2}x}dx=-\log|\cos x|+\frac{1}{2}\log|\cos^{2}x-1/2|+c \]

on any interval \(I\) that does not contain points of the form \(x=\pi/2+k\pi\) or of the form \(x=\pi/4+k\pi/2\) with \(k\) an integer.

Solution

(d) With the substitution \(y=\sqrt{x}\), \(x,y>0\), we have \(x=y^{2}\), we write \(dx=2y\ dy\) and we obtain

\[ \int\frac{x+\sqrt{x}}{2+\sqrt{x}}dx=2\int\frac{y^{3}+y^{2}}{y+2}dy. \]

Integrating the rational function, we have

\[ \begin{array}{l}\ds2\int\frac{y^{3}+y^{2}}{y+2}dy=2\int(y^{2}-y+2)dy-8\int\frac{1}{y+2}dy\\ \\ \ds=\frac{2}{3}y^{3}-y^{2}+4y-8\log(y+2)+c.\end{array} \]

Substituting back \(y=\sqrt{x}\), we conclude

\[ \int\frac{x+\sqrt{x}}{2+\sqrt{x}}dx=\frac{2}{3}x\sqrt{x}-x+4\sqrt{x}-8\log(\sqrt{x}+2)+c \]

on any interval \(I\subset(0,+\infty)\).

Solution

(e) Using the rationalizing substitution \(y=\tan\frac{x}{2}\), \(dx=\frac{2}{1+y^{2}}dy\) and taking into account that \(\cos x=\frac{1-y^{2}}{1+y^{2}}\), we have

\[ \int\frac{1}{\cos x}dx=-2\int\frac{1}{y^{2}-1}dy. \]

Using partial fractions, we then have

\[ -2\int\frac{1}{y^{2}-1}dy=\int\left(\frac{1}{y+1}-\frac{1}{y-1}\right)dy=\log\left|\frac{y+1}{y-1}\right|+c. \]

Going back to \(y=\tan\frac{x}{2}\), we conclude

\[ \int\frac{1}{\cos x}dx=\log\left|\frac{\tan(x/2)+1}{\tan(x/2)-1}\right|+c \]

on any interval \(I\) that does not contain points of the form \(x=\pi/2+k\pi\) with \(k\) an integer.

Solution

(f) Using the rationalizing substitution \(x=2\sin y\), \(-\pi/2<y<\pi/2\), \(-2<x<2\), \(dx=2\cos y\ dy\), and taking into account also that \(\cos y>0\), we obtain

\[ \int\sqrt{4-x^{2}}dx=4\int\cos^{2}y\ dy. \]

From \(4\cos^{2}y=2+2\cos(2y)\), we then have

\[ 4\int\cos^{2}y\ dy=2y+\sin(2y)=2y+2\sin y\cos y+c. \]

Now using \(\sin y=x/2\), \(y=\arcsin(x/2)\), \(\cos y=\sqrt{1-\sin^{2}y}=\frac{1}{2}\sqrt{4-x^{2}}\), we conclude

\[ \int\sqrt{4-x^{2}}dx=2\arcsin(x/2)+(x/2)\sqrt{4-x^{2}}+c \]

on any interval \(I\subset(-2,2)\).

Exercise 6

Compute the following limits

\[ {\rm (a)}\ \lim_{x\to0^{+}}\frac{1}{x^{3}}\int_{0}^{x^{2}}\log(1+\sqrt{t})dt,~~~ {\rm (b)}\ \lim_{x\to0^{+}}\frac{1}{x^{4}}\int_{0}^{x^{2}}(1-\cos\sqrt{t})dt \]

\[ {\rm (c)}\ \lim_{k\to+\infty}\left(2\int_{4}^{4e}\frac{\log(kx)}{x}dx-\log(k(k+1))\right),~~~ {\rm (d)}\ \lim_{k\to+\infty}\left(\int_{1}^{3}\log(k(x+2))dx-3\log(k+1)\right) \]

Solution

(a) Setting \(f(x)=\int_{0}^{x^{2}}\log(1+\sqrt{t})dt\), the limit takes the indeterminate form \(\frac{0}{0}\) given by \(\lim_{x\to0^{+}}\frac{f(x)}{x^{3}}\). The hypotheses of L'Hopital's rule are satisfied, so we can move on to computing the limit

\[ \lim_{x\to0^{+}}\frac{f'(x)}{3x^{2}}. \]

We compute the derivative \(f'(x)\) using the Fundamental theorem of calculus and the chain rule (derivative of a composite function):

\[ f'(x)=2x\log(1+\sqrt{x^{2}})=2x\log(1+x),\ \ x>0. \]

In conclusion

\[ \begin{array}{l}\ds\lim_{x\to0^{+}}\frac{1}{x^{3}}\int_{0}^{x^{2}}\log(1+\sqrt{t})dt=\lim_{x\to0^{+}}\frac{2x\log(1+x)}{3x^{2}}\\ \\ \ds=\frac{2}{3}\lim_{x\to0^{+}}\frac{\log(1+x)}{x}=\frac{2}{3}. \end{array} \]

Solution

(b) Setting \(f(x)=\int_{0}^{x^{2}}(1-\cos\sqrt{t})dt\), the limit takes the indeterminate form \(\frac{0}{0}\) given by \(\lim_{x\to0^{+}}\frac{f(x)}{x^{4}}\). The hypotheses of L'Hopital's rule are satisfied, so we can move on to computing the limit

\[ \lim_{x\to0^{+}}\frac{f'(x)}{4x^{3}}. \]

We compute the derivative \(f'(x)\) using the Fundamental theorem of calculus and the chain rule (derivative of a composite function):

\[ f'(x)=2x(1-\cos\sqrt{x^{2}})=2x(1-\cos x),\ \ x>0. \]

In conclusion

\[ \begin{array}{l}\ds\lim_{x\to0^{+}}\frac{1}{x^{4}}\int_{0}^{x^{2}}(1-\cos\sqrt{t})dt=\lim_{x\to0^{+}}\frac{2x(1-\cos x)}{4x^{3}}\\ \\ \ds=\frac{1}{2}\lim_{x\to0^{+}}\frac{1-\cos x}{x^{2}}=\frac{1}{4}. \end{array} \]

Solution

(c) We compute

\[ \begin{array}{l}\ds2\int_{4}^{4e}\frac{\log(kx)}{x}dx=[\log^{2}(kx)]_{4}^{4e}=\log^{2}(4ke)-\log^{2}(4k)\\ \\ \ds=(\log(4ke)-\log(4k))(\log(4ke)+\log(4k))\\ \\ \ds=\log\frac{4ke}{4k}\log(16k^{2}e)=\log(16k^{2}e). \end{array} \]

It follows that

\[ \begin{array}{l}\ds\lim_{k\to+\infty}\left(2\int_{4}^{4e}\frac{\log(kx)}{x}dx-\log(k(k+1))\right) =\lim_{k\to+\infty}\log(16k^{2}e)-\log(k(k+1))\\ \\ \ds=\lim_{k\to+\infty}\log\frac{16k^{2}e}{k^{2}+k}= \log(16e)=1+\log16. \end{array} \]

Solution

(d) We compute

\[ \begin{array}{l}\ds\int_{1}^{3}\log(k(x+2))dx=[x\log(kx+2k)]_{1}^{3}-\int_{1}^{3}\frac{x}{x+2}dx\\ \\ \ds=3\log(5k)-\log(3k)-\int_{1}^{3}\left(1-\frac{2}{x+2}\right)dx\\ \\ \ds=3\log(5k)-\log(3k)-2+2\log\frac{5}{3}. \end{array} \]

It follows that

\[ \begin{array}{l}\ds\lim_{k\to+\infty}\left(\int_{1}^{3}\log(k(x+2))dx-3\log(k+1)\right)\\ \\ \ds=\lim_{k\to+\infty}3\log(5k)-\log(3k)-2+2\log\frac{5}{3}-3\log(k+1)\\ \\ \ds=\lim_{k\to+\infty}3\log\frac{5k}{k+1}-\log(3k)-2+2\log\frac{5}{3}=-\infty. \end{array} \]