Integration Techniques, multiple pages

6. Integration of irrational functions

Irrational functions are little harder to integrate. There are cases that need some specfic techniques in or order to be integrated.

The method used will depend on the contents of the integrand. We have already used most of these techniques and this part is a kind of summary plus some additions.

case 1: Exponents with fractions

This is the case where we have multiple irrationals combined by the regular operations:

Problem 34

Evaluate: $\int \frac{\sqrt{x}}{\sqrt[4]{x^{3}+1}} \, dx$

We have to re-write using exponents:

$\int \frac{\sqrt{x}}{\sqrt[4]{x^{3}+1}} \, dx}={\displaystyle \int \frac{x^{\frac{1}{2}}}{x^{\frac{3}{4}}+1} \, dx$

We have to find the commun denominator of all the fractions in exponents.

In this case we have only 2 fractions $\frac{1}{2}$ and $\frac{3}{4}$

The common denominator is 4 here.

Let $x=u^{4}$

$dx=4u^{3}du$

$x^{\frac{1}{2}}=u^{\frac{1}{2}\cdot 4}=u^{2}$

$x^{\frac{3}{4}}=u^{\frac{3}{4}\cdot 4}=u^{3}$

Now we get

$\int \frac{x^{\frac{1}{2}}}{x^{\frac{3}{4}}+1} \, dx}={\displaystyle 4\int \frac{u^{2}}{u^{3}+1}\, u^{3}du }={\displaystyle 4\int \frac{u^{5}}{u^{3}+1}\, du$

When we divide it:

$\frac{u^{5}}{u^{3}+1}=u^{2}- \frac{u^{5}}{u^{3}+1}$

$4\int \frac{u^{5}}{u^{3}+1}\, du}={\displaystyle 4 \int u^{2}\, du- 4\int \frac{u^{2}}{u^{3}+1}\, du$

$4 \int u^{2}\, du- 4\int \frac{u^{2}}{u^{3}+1}\, du}={\displaystyle 4 \int u^{2}\, du- \frac{4}{3}\int \frac{3u^{2}}{u^{3}+1}\, du$

$4 \int u^{2}\, du- \frac{4}{3}\int \frac{3u^{2}}{u^{3}+1}\, du}={\displaystyle \frac{4}{3}u^{3}- \frac{4}{3} \ln \left | u^{3} \right |+ C$

Now back to $x$ :

$\int \frac{\sqrt{x}}{\sqrt[4]{x^{3}+1}} \, dx}={\displaystyle \frac{4}{3}x^{\frac{3}{4}}- \frac{4}{3} \ln \left | x^{\frac{3}{4}}+1 \right |+ C$

Finally:

$\int \frac{\sqrt{x}}{\sqrt[4]{x^{3}+1}} \, dx}={\displaystyle \frac{4}{3}\sqrt[4]{x^{3}}- \frac{4}{3} \ln \left |\sqrt[4]{x^{3}}+1 \right |+ C$

Problem 35

Evaluate: $\int \frac{\sqrt[6]{x}+1}{\sqrt[6]{x^{7}}+\sqrt[4]{x^{5}}} \, dx$

We have to re-write using exponents:

$\int \frac{\sqrt[6]{x}+1}{\sqrt[6]{x^{7}}+\sqrt[4]{x^{5}}} \, dx}={\displaystyle \int \frac{x^{\frac{1}{6}}+1}{x^{\frac{7}{6}}+x^{\frac{5}{4} }} \, dx$

The fractions are: $\frac{1}{6}$ , $\frac{7}{6}$ and $\frac{5}{4}$

The common denominator is the LCM of 6 and 4. It is 12.

Let $x=u^{12} \Rightarrow dx=12u^{11}du$

$\int \frac{\sqrt[6]{x}+1}{\sqrt[6]{x^{7}}+\sqrt[4]{x^{5}}} \, dx}={\displaystyle \int \frac{u^{2}+1}{u^{14}+u^{15} } \cdot 12u^{11}\, du$

$\int \frac{u^{2}+1}{u^{14}+u^{13} } \cdot 12u^{11}\, du}={\displaystyle 12\int \frac{u^{2}+1}{u^{3}+u^{4} }\, du$

$12\int \frac{u^{2}+1}{u^{3}+u^{4} }\, du}={\displaystyle 12\int \frac{u^{2}+1}{u^{3}(u+1) }\, du$

Let:

$\frac{u^{2}+1}{u^{3}(u+1)} }={\displaystyle \frac{A}{u}+ \frac{B}{u^{2}}+ \frac{C}{u^{3}} + \frac{D}{u+1}$

$A(u^{3}+u^{2})+ B(u^{2}+u)+C(u+1)+D(u^{3})$

For $u=0$

$C=1$

The terms $u^{2}$

$Au^{2}+Bu^{2}=u^{2}$

$A+B=1$

$A=1-B$

The terms in $u$ :

$Bu+Cu=0u$

$B+C=0$

$B=-C=-1$

$A=1-B=1+1=2$

The terms in $u^{3}$

$Au^{3}+Du^{3}=0u^{3}$

$A+D=0$

$D=-A=-2$

We can now write:

$\frac{u^{2}+1}{u^{3}(u+1) }}={\displaystyle \frac{2}{u}- \frac{1}{u^{2}}+ \frac{1}{u^{3}} - \frac{2}{u+1 }$

(1) $\begin{equation*} \begin{split} \displaystyle 12\int \frac{u^{2}+1}{u^{3}(u+1) }\, du}&= {\displaystyle 12\int \frac{2du}{u}- 12\int \frac{du}{u^{2}}+ 12\int \frac{du}{u^{3}}-12\int \frac{2du}{u+1} }\\ &= {\displaystyle 24 \ln |u|+\frac{12}{u}-\frac{6}{u^{2}}-24 \ln |u+1| +C} \end{split} \end{equation*}$

Back to $x$

We know that $x=u^{12}\Rightarrow u=x^{\frac{1}{12}}$

(2) $\begin{equation*} \begin{split} \displaystyle \int \frac{\sqrt[6]{x}+1}{\sqrt[6]{x^{7}}+\sqrt[4]{x^{5}}} \, dx}&= {\displaystyle 24 \ln \left | x^{\frac{1}{12}} \right |+\frac{12}{x^{\frac{1}{12}}}-\frac{6}{(x^{\frac{1}{12}})^{2}}-24 \ln \left |x^{\frac{1}{12}}+1 \right | +C}\\ &= {\displaystyle 24 \ln \left |\sqrt[12]{x} \right |+\frac{12}{\sqrt[12]{x}}-\frac{6}{\sqrt[6]{x}}-24 \ln \left |\sqrt[12]{x}+1 \right | +C} \end{split} \end{equation*}$

Finally:

$\int \frac{\sqrt[6]{x}+1}{\sqrt[6]{x^{7}}+\sqrt[4]{x^{5}}} \, dx}={\displaystyle 24 \ln \left |\sqrt[12]{x} \right |+\frac{12}{\sqrt[12]{x}}-\frac{6}{\sqrt[6]{x}}-24 \ln \left |\sqrt[12]{x}+1 \right | +C$

$\int \frac{\sqrt[6]{x}+1}{\sqrt[6]{x^{7}}+\sqrt[4]{x^{5}}} \, dx}={\displaystyle 2 \ln \left | x \right |+\frac{12}{\sqrt[12]{x}}-\frac{6}{\sqrt[6]{x}}-24 \ln \left |\sqrt[12]{x}+1 \right | +C$

case 2: A single fraction with an exponent as fraction

$\int \sqrt[n]{\frac{ax+b}{cx+d}} \, dx$

In this situation we have to use the relation:

$\frac{ax+b}{cx+d} =u^{n}$

Problem 36

Evaluate: $\int \frac{1}{x}\sqrt{\frac{1-x}{x}} \, dx$

Let $u^{2}=\frac{1-x}{x}$

$1-x=u^{2}x$

$u^{2}x+x=1$

$(u^{2}+1)x=1$

$x=\frac{1}{u^{2}+1}$

$dx=\frac{-2u}{(u^{2}+1)^{2}}\, du$

We get:

(3) $\begin{equation*} \begin{split} \displaystyle \int \frac{1}{x}\sqrt{\frac{1-x}{x}} \, dx }&={\displaystyle \int (u^{2}+1)u \cdot \frac{-2u}{(u^{2}+1)^{2}} \, du}\\ &= {\displaystyle -2\int \frac{u^{2}}{u^{2}+1} \, du}\\ &={\displaystyle -2\int \frac{u^{2}+1-1}{u^{2}+1} \, du}\\ &={\displaystyle -2\int du+2 \int \frac{du}{u^{2}+1} \, du}\\ &=-2u+2\ \arctan u+C \end{split} \end{equation*}$

Bax to $x$

${\displaystyle \int \frac{1}{x}\sqrt{\frac{1-x}{x}} \, dx }=-2\sqrt{\frac{1-x}{x}}+2\arctan \sqrt{\frac{1-x}{x}}+C$

Finally:

${\displaystyle \int \frac{1}{x}\sqrt{\frac{1-x}{x}} \, dx }=-2\sqrt{\frac{1-x}{x}}+2\arctan \sqrt{\frac{1-x}{x}}+C$

Problem 37

Evaluate: $\int \frac {\sqrt{x+4}}{x} \, dx$

Let $u^{2}=x+4$

$x=u^{2}-4$

$dx=2u\, du$

We get:

(4) $\begin{equation*} \begin{split} \displaystyle \int \frac {\sqrt{x+4}}{x} \, dx }&={\displaystyle \int \frac{u}{u^{2}-4}\cdot 2u \, du}\\ &= {\displaystyle 2\int \frac{u^{2}}{u^{2}-4}\, du}\\ &={\displaystyle 2\int \frac{u^{2}-4+4}{u^{2}-4}\, du}\\ &={\displaystyle 2\int du+8 \int \frac{du}{u^{2}-4}\, du}\\ &=2u+\frac{8}{4}\ln \left |\frac{u-2}{u+2} \right | +C \end{split} \end{equation*}$

Bax to $x$

$\int \frac {\sqrt{x+4}}{x} \, dx }={\displaystyle 2\sqrt{x+4}+2\ln \left |\frac{\sqrt{x+4}-2}{\sqrt{x+4}+2} \right | +C$

Finally:

$\int \frac {\sqrt{x+4}}{x} \, dx }={\displaystyle 2\sqrt{x+4}+2\ln \left |\frac{\sqrt{x+4}-2}{\sqrt{x+4}+2} \right | +C$

case 3: Irrational functions in denominator with polynomials in numerator

This form has been solved already :

The form is:

$\int \frac{ax+b}{\sqrt{ax^{2}+bx +c}}\, dx$

Problem 38

Evaluate: $\int \frac{4x-1}{\sqrt{5-4x-x^{2}}} \, dx$

We know that:

${\displaystyle \frac{d(5-4x-x^{2})}{dx} }=-4-2x=-2x-4$

Also:

$-(2+x)^{2}=-(4+4x+x^{2})=-4-4x-x^{2}$

$5-4x-x^{2}=9-(4+4x+x^{2}=9-(2+x)^{2}=3^{2}-(2+x)^{2}$

From $4x-1$

$4x-1=-2(-2x-4)-9$

Back to the evaluation:

(5) $\begin{equation*} \begin{split} \displaystyle \int \frac{4x-1}{\sqrt{5-4x-x^{2}}} \, dx }&={\displaystyle -2 \int \frac{-2x-4}{\sqrt{5-4x-x^{2}}} \, dx -9 \int \frac{dx}{\sqrt{5-4x-x^{2}}} }\\ &= {\displaystyle -2 \int \frac{-2x-4}{\sqrt{5-4x-x^{2}}} \, dx -9 \int \frac{dx}{\sqrt{3^{2}-(2+x)^{2}}} }\\ &={\displaystyle -4\sqrt{5-4x-x^{2}}-9\sin^{-1} \frac{2+x}{3} +C} \end{split} \end{equation*}$