What is the integral of 9x^2sqrt(x^3+1) ?

The integral of 9x^2sqrt(x^3+1) is 2(x^3+1)^{3/2}+C

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integral of 9x^2sqrt(x^3+1)

\int 9 x^{2} x^{3} + 1 d x

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

Solution steps

\int 9 x^{2} x^{3} + 1 d x

Take the constant out:

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

= 9 \cdot \int x^{2} x^{3} + 1 d x

Apply u-substitution

: \int \frac{u}{3} d u

= 9 \cdot \int \frac{u}{3} d u

Take the constant out:

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

= 9 \cdot \frac{1}{3} \cdot \int u d u

Apply the Power Rule

: \frac{2}{3} u^{\frac{3}{2}}

= 9 \cdot \frac{1}{3} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = x^{3} + 1

= 9 \cdot \frac{1}{3} \cdot \frac{2}{3} (x^{3} + 1)^{\frac{3}{2}}

Simplify

9 \cdot \frac{1}{3} \cdot \frac{2}{3} (x^{3} + 1)^{\frac{3}{2}} : 2 (x^{3} + 1)^{\frac{3}{2}}

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

Add a constant to the solution

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

\int \frac{4 x ^{2}}{( x ^{2} - 9 ) ^{\frac{3}{2}}} d x

\int \frac{e ^{- x}}{e ^{- 2 x} + 1} d x

\int 2 (2 x + 3)^{3} d x

\int \frac{x ^{2} - 2 x + 3}{x} d x

What is the integral of 9x^2sqrt(x^3+1) ?
The integral of 9x^2sqrt(x^3+1) is 2(x^3+1)^{3/2}+C