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

The integral of (5x^3+1)^4x^5 is (625x^{18}}{18}+\frac{100x^{15})/3+(25x^{12})/2+(20x^9)/9+(x^6)/6+C

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integral of (5x^3+1)^4x^5

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

\frac{625 x ^{18}}{18} + \frac{100 x ^{15}}{3} + \frac{25 x ^{12}}{2} + \frac{20 x ^{9}}{9} + \frac{x ^{6}}{6} + C

Solution steps

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

Expand

(5 x^{3} + 1)^{4} x^{5} : 625 x^{17} + 500 x^{14} + 150 x^{11} + 20 x^{8} + x^{5}

= \int 625 x^{17} + 500 x^{14} + 150 x^{11} + 20 x^{8} + x^{5} 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^{17} d x + \int 500 x^{14} d x + \int 150 x^{11} d x + \int 20 x^{8} d x + \int x^{5} d x

\int 625 x^{17} d x = \frac{625 x ^{18}}{18}

\int 500 x^{14} d x = \frac{100 x ^{15}}{3}

\int 150 x^{11} d x = \frac{25 x ^{12}}{2}

\int 20 x^{8} d x = \frac{20 x ^{9}}{9}

\int x^{5} d x = \frac{x ^{6}}{6}

= \frac{625 x ^{18}}{18} + \frac{100 x ^{15}}{3} + \frac{25 x ^{12}}{2} + \frac{20 x ^{9}}{9} + \frac{x ^{6}}{6}

Add a constant to the solution

= \frac{625 x ^{18}}{18} + \frac{100 x ^{15}}{3} + \frac{25 x ^{12}}{2} + \frac{20 x ^{9}}{9} + \frac{x ^{6}}{6} + C

What is the integral of (5x^3+1)^4x^5 ?
The integral of (5x^3+1)^4x^5 is (625x^{18}}{18}+\frac{100x^{15})/3+(25x^{12})/2+(20x^9)/9+(x^6)/6+C