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.

Type 1
Integrand contains $$\sqrt[n]{\frac{ax+b}{cx+d}}$$

Use the substitution $$u=\sqrt[n]{\frac{ax+b}{cx+d}}$$. Find $$\int\frac{1}{x}\sqrt{\frac{1-x}{x}}\,dx$$.
 * Example

Find $$\int\frac{x}{\sqrt[3]{ax+b}}\,dx$$.

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

Write $$Px+Q$$ as $$Px+Q=p\cdot\frac{d[ax^2+bx+c]}{dx}+q$$. Find $$\int\frac{4x-1}{\sqrt{5-4x-x^2}}\,dx$$.
 * Example

Type 3
Integrand contains $$\sqrt{a^2-x^2}$$, $$\sqrt{a^2+x^2}$$ or $$\sqrt{x^2-a^2}$$

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


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

Type 4
Integral is of the form $$\int\frac{dx}{(px+q)\sqrt{ax^2+bx+c}}$$

Use the substitution $$u=\frac{1}{px+q}$$. Find $$\int\frac{dx}{(1+x)\sqrt{3+6x+x^2}}$$.
 * Example

Type 5
Other rational expressions with the irrational function $$\sqrt{ax^2+bx+c}$$


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