What is the integral of (3x^2+6x)(x^3+3x^2+1)^{2/3} ?

The integral of (3x^2+6x)(x^3+3x^2+1)^{2/3} is 3/5 (x^3+3x^2+1)^{5/3}+C

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integral of (3x^2+6x)(x^3+3x^2+1)^{2/3}

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

\frac{3}{5} (x^{3} + 3 x^{2} + 1)^{\frac{5}{3}} + C

Solution steps

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

Apply u-substitution

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

= \int u^{\frac{2}{3}} d u

Apply the Power Rule

: \frac{3}{5} u^{\frac{5}{3}}

= \frac{3}{5} u^{\frac{5}{3}}

Substitute back

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

= \frac{3}{5} (x^{3} + 3 x^{2} + 1)^{\frac{5}{3}}

Add a constant to the solution

= \frac{3}{5} (x^{3} + 3 x^{2} + 1)^{\frac{5}{3}} + C

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What is the integral of (3x^2+6x)(x^3+3x^2+1)^{2/3} ?
The integral of (3x^2+6x)(x^3+3x^2+1)^{2/3} is 3/5 (x^3+3x^2+1)^{5/3}+C