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

The integral of (x^4-5x)^2(4x^3-5) is (x^{12})/3-5x^9+25x^6-(125x^3)/3+C

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

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

\frac{x ^{12}}{3} - 5 x^{9} + 25 x^{6} - \frac{125 x ^{3}}{3} + C

Solution steps

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

Expand

(x^{4} - 5 x)^{2} (4 x^{3} - 5) : 4 x^{11} - 45 x^{8} + 150 x^{5} - 125 x^{2}

= \int 4 x^{11} - 45 x^{8} + 150 x^{5} - 125 x^{2} 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 4 x^{11} d x - \int 45 x^{8} d x + \int 150 x^{5} d x - \int 125 x^{2} d x

\int 4 x^{11} d x = \frac{x ^{12}}{3}

\int 45 x^{8} d x = 5 x^{9}

\int 150 x^{5} d x = 25 x^{6}

\int 125 x^{2} d x = \frac{125 x ^{3}}{3}

= \frac{x ^{12}}{3} - 5 x^{9} + 25 x^{6} - \frac{125 x ^{3}}{3}

Add a constant to the solution

= \frac{x ^{12}}{3} - 5 x^{9} + 25 x^{6} - \frac{125 x ^{3}}{3} + C

\int \frac{2 x}{2 - x ^{2}} d x

\int \frac{3 x ^{3}}{x ^{2} + x - 2} d x

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

\int \frac{x ^{4} + 1}{x ^{4}} d x

\int \frac{1}{1 - x} d x

What is the integral of (x^4-5x)^2(4x^3-5) ?
The integral of (x^4-5x)^2(4x^3-5) is (x^{12})/3-5x^9+25x^6-(125x^3)/3+C