5.5 Basic Integration Techniques

5.5.1 Rational Functions with Linear Denominators

Here we evaluate integrals of the form
\[\int \frac{P(x)}{ax + b}\,dx, \;\; a\neq 0, \;\; \text{and } \; P(x) \; \text{ a polynomial}.\]
The following points must be kept in mind:

1. In the special case where \(P(x)=1\), by the chain rule, we have
   \[\int \frac{1}{ax + b}\,dx = \frac{1}{a}\int \frac{a}{ax + b}\,dx = \frac{1}{a}\ln|ax+b| +C.\]
2. If (deg. of num.) \(\geq\) (deg. of denom.), use polynomial division to rewrite the rational expression.

Example 1. Evaluate \(\displaystyle \int \frac{4x^2+6x-3}{2x+1} \, dx\).

Solution. Since (degree of the numerator) \(\geq\) (degree of the denominator), then
\[\frac{4x^2+6x-3}{2x+1}=2+2 x-\frac{5}{2 x+1} \Longrightarrow \int \frac{4x^2+6x-3}{2x+1} \, dx =\int \left(2+2 x-\frac{5}{2x+1} \right)dx.\]
Then, using (1) above,
\[ \int \frac{4x^2+6x-3}{2x+1} \, dx= \int \left(2+2 x-\frac{5}{2x+1} \right) \, dx = 2x+x^2-\frac{5}{2}\ln |2x+1|+C.\]

5.5.3 Algebraic Manipulation

Example 1. Evaluate \(\displaystyle \int \frac{x}{\sqrt{2x+3}} \, dx\).

Solution. Let \(u=2x+3, d u = 2\,d x\), then \(\displaystyle x=\frac{u-3}{2}\), and

\begin{array}{rl}
\displaystyle \int \frac{x}{\sqrt{2x+3}} \, dx
&=
\displaystyle \frac{1}{2}\int \frac{\frac{u-3}{2}}{\sqrt{u}} \, du=\frac{1}{4}\int \frac{u-3}{\sqrt{u}} \, du =\frac{1}{4}\int
\frac{u-3}{u^{1/2}} \, du=\frac{1}{4}\int \!\left(\!u^{1/2}-3u^{-1/2}\right) du = \\
&=
\displaystyle \frac{1}{4}\left(\frac{u^{3/2}}{3/2}-3\frac{u^{1/2}}{1/2}\right)+C=\frac{1}{4}\left(\frac{2u\sqrt{u}}{3}-6\sqrt{u}\right)+C=\frac{2\sqrt{u}}{4}\!\left(\frac{u}{3}-3\!\right)+C =\\
&=
\displaystyle \frac{2\sqrt{2x+3}}{4}\left(\frac{2x+3-9}{3}\right)+C\underset{?}{=}\frac{1}{3}(x-3)\sqrt{2x+3}+C.
\end{array}

Example 2. Evaluate \(\displaystyle \int \frac{x}{\left(1+\sqrt{x}\right)^2} \, dx\)

Solution. Let \(u= 1+\sqrt{x}\), then \(\displaystyle d u = \frac{1}{2}\frac{1}{\sqrt{x}}\,d x\). It follows that
\[u= 1+\sqrt{x} \Longrightarrow \sqrt{x}= u-1 \Longrightarrow x = (u-1)^2,\]
\[ d u = \frac{1}{2}\frac{1}{\sqrt{x}}\,d x \Longrightarrow dx =2\sqrt{x}\,d u = 2(u-1)\,d u.\]
Rewriting the original integral in terms of \(u\),
\[
\renewcommand{\arraystretch}{2}
\begin{array}{rl}
\displaystyle \int \frac{x}{\left(1+\sqrt{x}\right)^2} \, dx
&=\displaystyle \int \frac{(u-1)^2}{u^2}2(u-1)du =2\int \frac{(u-1)^3}{u^2} \, du = 2\int \frac{u^3-3u^2+3u-1}{u^2}
\, du \\
&=\displaystyle 2\int \frac{u^3-3u^2+3u-1}{u^2} \, du= 2\int \left(u -3 +\frac{3}{u}-\frac{1}{u^2}\right)\, du =\\
&=\displaystyle 2\left(\frac{1}{2}u^2-3u+3\ln u+\frac{1}{u}\right)+C
= u^2-6u+6\ln u+\frac{2}{u}+C = \\
&=\displaystyle \left(1+\sqrt{x}\right)^2-6\left(1+\sqrt{x}\right)+6\ln \left(1+\sqrt{x}\right)+\frac{2}{1+\sqrt{x}}+C=\\
&=\displaystyle 1+2\sqrt{x} +x-6 -6\sqrt{x}+6\ln \left(1+\sqrt{x}\right)+\frac{2}{1+\sqrt{x}}+C =\\
&\displaystyle \underset{?}{=} x-4\sqrt{x} +6\ln \left(1+\sqrt{x}\right)+\frac{2}{1+\sqrt{x}}+K.
\end{array}.
\]

Example 3. Evaluate \(\displaystyle \int \frac{x^7}{(1+x^4)^3} \, dx\).

Solution. Set \(u=1+x^4\), \(d u = 4x^3\,d x\). Then \(x^4=u-1\), and

\begin{array}{rl}
\displaystyle \int \frac{x^7}{(1+x^4)^3} \, dx
&=\displaystyle \frac{1}{4}\int \frac{x^4}{(1+x^4)^3} 4 x^3\, dx
=\frac{1}{4}\int \frac{u-1}{u^3} \, du
=\frac{1}{4}\int \left( \frac{1}{u^2}-\frac{1}{u^3}\right) \, du =\\
&\displaystyle =\frac{1}{4}\int \left( u^{-2}-u^{-3}\right) \, du
=\frac{1}{4}\left( -\frac{1}{u}+\frac{1}{2}\frac{1}{u^2} \right) +C
=-\frac{1}{8}\left( \frac{2u-1}{u^2} \right) +C =\\
& \displaystyle=-\frac{1}{8}\left( \frac{2(1+x^4)-1}{(1+x^4)^2} \right) +C
=-\frac{1}{8}\frac{2x^4+1}{(1+x^4)^2} + C.
\end{array}

5.5.5 Inverse Trigonometric Function

Recall
\[ \int \frac{1}{\sqrt{1-x^2}}\,d x=\arcsin x,\;\;
\int \frac{1}{1+x^2} \,d x=\arctan x+C,\;\;
\int \frac{1}{x\sqrt{x^2-1}}\, d x=\text{arcsec}\, x+C,\; x>1.\]

Example 1. Evaluate (a) \(\displaystyle \int \frac{1}{x\sqrt{2x^2-1}} \, dx\), \(\displaystyle x>\frac{1}{\sqrt{2}}\);    (b) \(\displaystyle \int \frac{1}{\sqrt{8-x^2}} \, dx\).

Solution. (a) This integral is similar to the indefinite integral for \(\text{arcsec}\, x\). Thus, we can rewrite it as follows:
\[
\displaystyle \int \frac{1}{x\sqrt{2x^2-1}} \, dx
=\large{\int} \frac{1}{x\sqrt{\left(\sqrt{2}\,x\right)^2-1}} \, dx..\]
Thus, set \(u=\sqrt{2}\,x\), \(d u =\sqrt{2}\,d x\). Then
\begin{align*}
\large{\int} \frac{1}{x\sqrt{\left(\sqrt{2}\,x-1\right)^2}} \, dx
&=\large{\int} \frac{1}{\sqrt{2}\,x\sqrt{\left(\sqrt{2}\,x\right)^2-1}}\sqrt{2}\,dx
=\int \frac{1}{u\sqrt{u^2-1}}\,d u=\text{arcsec}\, u+C =\\
&= \text{arcsec}\, \sqrt{2}\,x+C.
\end{align*}

(b) This integral is similar to the indefinite integral for \(\arcsin x\), thus factor the 8 inside the radical.
\[
\int \frac{1}{\sqrt{8-x^2}} \, dx
=\displaystyle \large{\int} \frac{1}{\displaystyle \sqrt{8\left(1- x^2/8\right)}} \, dx
=\large{\int} \frac{1}{\sqrt{8}}\frac{1}{\displaystyle \sqrt{1-\left(x/\sqrt{8}\right)^2}}
\, dx.\]
Let \(\displaystyle u = \frac{x}{\sqrt{8}}\), then \(\displaystyle d u = \frac{1}{\sqrt{8}}\,d x \). And
\[\int \frac{1}{\sqrt{8-x^2}} \, dx
= \int \frac{1}{\sqrt{1-u^2}} \, du
= \arcsin u + C = \arcsin \frac{x}{2\sqrt{2}}+C.
\]

Example 2. Evaluate (a) \(\displaystyle \int \frac{1}{x^2+4x+6} \, dx\);     (b) \(\displaystyle \int \frac{1}{\sqrt{-2 x^2 + 4 x + 6}} \, dx\).

Solution. (a) Because of the linear term, first complete the square.
\[
\int \frac{1}{x^2+4x+7}\, dx
=\int\frac{1}{(x+2)^2+3} \, dx
=\large{\int} \frac{1}{3\left(\frac{1}{3}(x+2)^2+1\right)} \, dx
=\large{\int} \frac{1}{3\left(\left(\frac{x+2}{\sqrt{3}}\right)^2+1\right)} \, dx.\]
Thus, let \(\displaystyle u=\frac{x+2}{\sqrt{3}}\), then \(\displaystyle d u =\frac{1}{\sqrt{3}}\,d x\). And
\[
\large{\int} \frac{1}{3\left(\left(\frac{x+2}{\sqrt{3}}\right)^2+1\right)} \, dx
=\large{\int} \frac{\sqrt{3}}{3}\frac{1}{u^2+1}\,du
=\frac{\sqrt{3}}{3}\arctan u+C
=\frac{\sqrt{3}}{3}\arctan \frac{x+2}{\sqrt{3}}+C.\]

(b) As in (a), first complete the square.
\[
\renewcommand{\arraystretch}{2.5}
\begin{array}{rl}
\displaystyle \large{\int} \frac{1}{\sqrt{-2 x^2 + 4 x + 6}} \, dx
&\displaystyle =\large{\int} \frac{1}{\sqrt{-2(x^2-2x)+6}} \, dx
=\large{\int} \frac{1}{\sqrt{-2(x^2-2x+1)+2+6}} \, dx
= \\
&\displaystyle =\large{\int} \frac{1}{\sqrt{8-2(x-1)^2}} \, dx
\underset{?}{=}\large{\int} \frac{1}{2\sqrt{2}\sqrt{1-\left(\frac{x-1}{2}\right)^2}} \, dx.\end{array}\]
Let \(\displaystyle u = \frac{x-1}{2} \), then \(\displaystyle d u = \frac{1}{2}\,d x \).
\[\large{\int} \frac{1}{\sqrt{-2 x^2 + 4 x + 6}} \, dx
\underset{?}{=}\large{\int}\frac{1}{\sqrt{2}} \frac{1}{\sqrt{1-u^2}} \, du
= \frac{1}{\sqrt{2}}\arcsin u + C = \frac{1}{\sqrt{2}}\arcsin \frac{x-1}{2} +C.
\]