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

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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

= \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

: \int \frac{v}{a} d v

= \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{2}{3} v^{\frac{3}{2}}

= \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

r x + d

\frac{\partial}{\partial x} (4 y^{3} sin (5 x))

\frac{s ^{2} - s + 1}{( s + 1 ) ^{2}}

x^{2} - 5, 4

x y^{^{'}} - y = 2 x^{2}

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