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

The integral of (2x+3)(x^2+3x)^2 is (2x^6)/5-(x^6}{15}+3x^5+6x^4+\frac{9x^4)/2-(3x^4)/2+9x^3+C

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integral of (2x+3)(x^2+3x)^2

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

\frac{2 x ^{6}}{5} - \frac{x ^{6}}{15} + 3 x^{5} + 6 x^{4} + \frac{9 x ^{4}}{2} - \frac{3 x ^{4}}{2} + 9 x^{3} + C

Solution steps

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

Apply Integration By Parts

: \frac{2 x ^{6}}{5} + 3 x^{5} + 6 x^{4} + \frac{3 x ^{5}}{5} + \frac{9 x ^{4}}{2} + 9 x^{3} - \int \frac{2 x ^{5}}{5} + 3 x^{4} + 6 x^{3} d x

= \frac{2 x ^{6}}{5} + 3 x^{5} + 6 x^{4} + \frac{3 x ^{5}}{5} + \frac{9 x ^{4}}{2} + 9 x^{3} - \int \frac{2 x ^{5}}{5} + 3 x^{4} + 6 x^{3} d x

\int \frac{2 x ^{5}}{5} + 3 x^{4} + 6 x^{3} d x = \frac{x ^{6}}{15} + \frac{3 x ^{5}}{5} + \frac{3 x ^{4}}{2}

= \frac{2 x ^{6}}{5} + 3 x^{5} + 6 x^{4} + \frac{3 x ^{5}}{5} + \frac{9 x ^{4}}{2} + 9 x^{3} - (\frac{x ^{6}}{15} + \frac{3 x ^{5}}{5} + \frac{3 x ^{4}}{2})

Simplify

= \frac{2 x ^{6}}{5} - \frac{x ^{6}}{15} + 3 x^{5} + 6 x^{4} + \frac{9 x ^{4}}{2} - \frac{3 x ^{4}}{2} + 9 x^{3}

Add a constant to the solution

= \frac{2 x ^{6}}{5} - \frac{x ^{6}}{15} + 3 x^{5} + 6 x^{4} + \frac{9 x ^{4}}{2} - \frac{3 x ^{4}}{2} + 9 x^{3} + C

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What is the integral of (2x+3)(x^2+3x)^2 ?
The integral of (2x+3)(x^2+3x)^2 is (2x^6)/5-(x^6}{15}+3x^5+6x^4+\frac{9x^4)/2-(3x^4)/2+9x^3+C