What is the integral of (3/5 x^8-4/3 x^3)^2 ?

The integral of (3/5 x^8-4/3 x^3)^2 is (9x^{17})/(425)-(2x^{12})/(15)+(16x^7)/(63)+C

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integral of (3/5 x^8-4/3 x^3)^2

\int (\frac{3}{5} x^{8} - \frac{4}{3} x^{3})^{2} d x

\frac{9 x ^{17}}{425} - \frac{2 x ^{12}}{15} + \frac{16 x ^{7}}{63} + C

Solution steps

\int (\frac{3}{5} x^{8} - \frac{4}{3} x^{3})^{2} d x

Simplify

(\frac{3}{5} x^{8} - \frac{4}{3} x^{3})^{2} : \frac{9}{25} x^{16} - \frac{8 x ^{11}}{5} + \frac{16}{9} x^{6}

= \int \frac{9}{25} x^{16} - \frac{8 x ^{11}}{5} + \frac{16}{9} x^{6} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{9}{25} x^{16} d x - \int \frac{8 x ^{11}}{5} d x + \int \frac{16}{9} x^{6} d x

\int \frac{9}{25} x^{16} d x = \frac{9 x ^{17}}{425}

\int \frac{8 x ^{11}}{5} d x = \frac{2 x ^{12}}{15}

\int \frac{16}{9} x^{6} d x = \frac{16 x ^{7}}{63}

= \frac{9 x ^{17}}{425} - \frac{2 x ^{12}}{15} + \frac{16 x ^{7}}{63}

Add a constant to the solution

= \frac{9 x ^{17}}{425} - \frac{2 x ^{12}}{15} + \frac{16 x ^{7}}{63} + C

What is the integral of (3/5 x^8-4/3 x^3)^2 ?
The integral of (3/5 x^8-4/3 x^3)^2 is (9x^{17})/(425)-(2x^{12})/(15)+(16x^7)/(63)+C