What is the integral of x^nsqrt(ax^{n+1)+b} ?

The integral of x^nsqrt(ax^{n+1)+b} is (2(ax^{n+1}+b)^{3/2})/(3a(n+1))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of x^nsqrt(ax^{n+1)+b}

\int x^{n} a x^{n + 1} + b d x

\frac{2 ( a x ^{n + 1} + b ) ^{\frac{3}{2}}}{3 a ( n + 1 )} + C

Solution steps

\int x^{n} a x^{n + 1} + b d x

Apply u-substitution

= \int \frac{a u + b}{n + 1} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{n + 1} \cdot \int a u + b d u

Apply u-substitution

= \frac{1}{n + 1} \cdot \int \frac{v}{a} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{n + 1} \cdot \frac{1}{a} \cdot \int v d v

Apply the Power Rule

= \frac{1}{n + 1} \cdot \frac{1}{a} \cdot \frac{2}{3} v^{\frac{3}{2}}

Substitute back

= \frac{1}{n + 1} \cdot \frac{1}{a} \cdot \frac{2}{3} (a x^{n + 1} + b)^{\frac{3}{2}}

Simplify

\frac{1}{n + 1} \cdot \frac{1}{a} \cdot \frac{2}{3} (a x^{n + 1} + b)^{\frac{3}{2}} : \frac{2 ( a x ^{n + 1} + b ) ^{\frac{3}{2}}}{3 a ( n + 1 )}

= \frac{2 ( a x ^{n + 1} + b ) ^{\frac{3}{2}}}{3 a ( n + 1 )}

Add a constant to the solution

= \frac{2 ( a x ^{n + 1} + b ) ^{\frac{3}{2}}}{3 a ( n + 1 )} + C

What is the integral of x^nsqrt(ax^{n+1)+b} ?
The integral of x^nsqrt(ax^{n+1)+b} is (2(ax^{n+1}+b)^{3/2})/(3a(n+1))+C