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

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

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

\int - 4 sin (- 2 t) d t

\int 3 t^{2} (t^{3} + 4)^{5} d t

\frac{1}{1 + \frac{1}{x}}, 0

\int_{- 1}^{1} (x - 1 - x^{2}) d x

t y^{^{'}} = 3 y + t^{4} sin (4 t)

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