What is the integral of (x^2+4x)sqrt(x^3+6x^2+1) ?

The integral of (x^2+4x)sqrt(x^3+6x^2+1) is 2/9 (x^3+6x^2+1)^{3/2}+C

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integral of (x^2+4x)sqrt(x^3+6x^2+1)

\int (x^{2} + 4 x) x^{3} + 6 x^{2} + 1 d x

\frac{2}{9} (x^{3} + 6 x^{2} + 1)^{\frac{3}{2}} + C

Solution steps

\int (x^{2} + 4 x) x^{3} + 6 x^{2} + 1 d x

Apply u-substitution

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

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

Take the constant out:

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

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

Apply the Power Rule

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

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

Substitute back

u = x^{3} + 6 x^{2} + 1

= \frac{1}{3} \cdot \frac{2}{3} (x^{3} + 6 x^{2} + 1)^{\frac{3}{2}}

Simplify

\frac{1}{3} \cdot \frac{2}{3} (x^{3} + 6 x^{2} + 1)^{\frac{3}{2}} : \frac{2}{9} (x^{3} + 6 x^{2} + 1)^{\frac{3}{2}}

= \frac{2}{9} (x^{3} + 6 x^{2} + 1)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{9} (x^{3} + 6 x^{2} + 1)^{\frac{3}{2}} + C

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What is the integral of (x^2+4x)sqrt(x^3+6x^2+1) ?
The integral of (x^2+4x)sqrt(x^3+6x^2+1) is 2/9 (x^3+6x^2+1)^{3/2}+C