Calculus/Integration techniques/Irrational Functions

Integration of irrational functions is more difficult than rational functions, and many cannot be done. However, there are some particular types that can be reduced to rational forms by suitable substitutions.

Contents

Type 1Edit

Integrand contains ${\sqrt[{n}]{\frac {ax+b}{cx+d}}}$${\sqrt[ {n}]{{\frac {ax+b}{cx+d}}}}$

Use the substitution $u={\sqrt[{n}]{\frac {ax+b}{cx+d}}}$$u={\sqrt[ {n}]{{\frac {ax+b}{cx+d}}}}$ .

Example

Find $\int {\frac {1}{x}}{\sqrt {\frac {1-x}{x}}}\,dx$$\int {\frac {1}{x}}{\sqrt {\frac {1-x}{x}}}\,dx$ .

Failed to parse (syntax error): {\displaystyle \int\frac{x}{\sqrt[3]{ax+b}}\,dx\end{align}}

Type 2Edit

Integral is of the form $\int {\frac {Px+Q}{\sqrt {ax^{2}+bx+c}}}\,dx$$\int {\frac {Px+Q}{\sqrt {ax^{2}+bx+c}}}\,dx$

Write $Px+Q$$Px+Q$ as $Px+Q=p\cdot {\frac {d[ax^{2}+bx+c]}{dx}}+q$$Px+Q=p\cdot {\frac {d[ax^{2}+bx+c]}{dx}}+q$ .

Example

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

Type 3Edit

Integrand contains ${\sqrt {a^{2}-x^{2}}}$${\sqrt {a^{2}-x^{2}}}$ , ${\sqrt {a^{2}+x^{2}}}$${\sqrt {a^{2}+x^{2}}}$ or ${\sqrt {x^{2}-a^{2}}}$${\sqrt {x^{2}-a^{2}}}$

This was discussed in "trigonometric substitutions above". Here is a summary:

1. For ${\sqrt {a^{2}-x^{2}}}$${\sqrt {a^{2}-x^{2}}}$ , use $x=a\sin(\theta )$$x=a\sin(\theta )$ .
2. For ${\sqrt {a^{2}+x^{2}}}$${\sqrt {a^{2}+x^{2}}}$ , use $x=a\tan(\theta )$$x=a\tan(\theta )$ .
3. For ${\sqrt {x^{2}-a^{2}}}$${\sqrt {x^{2}-a^{2}}}$ , use $x=a\sec(\theta )$$x=a\sec(\theta )$ .

Type 4Edit

Integral is of the form $\int {\frac {dx}{(px+q){\sqrt {ax^{2}+bx+c}}}}$$\int {\frac {dx}{(px+q){\sqrt {ax^{2}+bx+c}}}}$

Use the substitution $u={\frac {1}{px+q}}$$u={\frac {1}{px+q}}$ .

Example

Find $\int {\frac {dx}{(1+x){\sqrt {3+6x+x^{2}}}}}$$\int {\frac {dx}{(1+x){\sqrt {3+6x+x^{2}}}}}$ .

Type 5Edit

Other rational expressions with the irrational function ${\sqrt {ax^{2}+bx+c}}$${\sqrt {ax^{2}+bx+c}}$

1. If $a>0$$a>0$ , we can use $u={\sqrt {ax^{2}+bx+c}}\pm {\sqrt {a}}x$$u={\sqrt {ax^{2}+bx+c}}\pm {\sqrt {a}}x$ .
2. If $c>0$$c>0$ , we can use $u={\frac {{\sqrt {ax^{2}+bx+c}}\pm {\sqrt {c}}}{x}}$$u={\frac {{\sqrt {ax^{2}+bx+c}}\pm {\sqrt {c}}}{x}}$ .
3. If $ax^{2}+bx+c$$ax^{2}+bx+c$ can be factored as $a(x-\alpha )(x-\beta )$$a(x-\alpha )(x-\beta )$ , we can use $u={\sqrt {\frac {a(x-\alpha )}{x-\beta }}}$$u={\sqrt {{\frac {a(x-\alpha )}{x-\beta }}}}$ .
4. If $a<0$$a<0$ and $ax^{2}+bx+c$$ax^{2}+bx+c$ can be factored as $-a(\alpha -x)(x-\beta )$$-a(\alpha -x)(x-\beta )$ , we can use $x=\alpha \cos ^{2}(\theta )+\beta \sin ^{2}(\theta )$$x=\alpha \cos ^{2}(\theta )+\beta \sin ^{2}(\theta )$