What is the integral of x^2(5x^3-1)^4 ?

The integral of x^2(5x^3-1)^4 is (125x^{15})/3-(125x^{12})/3+(50x^9)/3-(10x^6)/3+(x^3)/3+C

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integral of x^2(5x^3-1)^4

\int x^{2} (5 x^{3} - 1)^{4} d x

\frac{125 x ^{15}}{3} - \frac{125 x ^{12}}{3} + \frac{50 x ^{9}}{3} - \frac{10 x ^{6}}{3} + \frac{x ^{3}}{3} + C

Solution steps

\int x^{2} (5 x^{3} - 1)^{4} d x

Expand

x^{2} (5 x^{3} - 1)^{4} : 625 x^{14} - 500 x^{11} + 150 x^{8} - 20 x^{5} + x^{2}

= \int 625 x^{14} - 500 x^{11} + 150 x^{8} - 20 x^{5} + 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 625 x^{14} d x - \int 500 x^{11} d x + \int 150 x^{8} d x - \int 20 x^{5} d x + \int x^{2} d x

\int 625 x^{14} d x = \frac{125 x ^{15}}{3}

\int 500 x^{11} d x = \frac{125 x ^{12}}{3}

\int 150 x^{8} d x = \frac{50 x ^{9}}{3}

\int 20 x^{5} d x = \frac{10 x ^{6}}{3}

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

= \frac{125 x ^{15}}{3} - \frac{125 x ^{12}}{3} + \frac{50 x ^{9}}{3} - \frac{10 x ^{6}}{3} + \frac{x ^{3}}{3}

Add a constant to the solution

= \frac{125 x ^{15}}{3} - \frac{125 x ^{12}}{3} + \frac{50 x ^{9}}{3} - \frac{10 x ^{6}}{3} + \frac{x ^{3}}{3} + C

\int \frac{x + 1}{x ^{2} - 16} d x

n = 0 \sum \infty 4 n^{2} + 4 n + 1

y = 4 x^{2}, y = x^{2} + 2

x \to 0 lim ((1 - 4 x)^{\frac{1}{x}})

x y^{^{'}} + x y = 1 - y, y (1) = 0

What is the integral of x^2(5x^3-1)^4 ?
The integral of x^2(5x^3-1)^4 is (125x^{15})/3-(125x^{12})/3+(50x^9)/3-(10x^6)/3+(x^3)/3+C