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

The integral of (x^4+5)^3x^3 is (x^{16}}{16}+\frac{5x^{12})/4+(75x^8)/8+(125x^4)/4+C

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

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

\frac{x ^{16}}{16} + \frac{5 x ^{12}}{4} + \frac{75 x ^{8}}{8} + \frac{125 x ^{4}}{4} + C

Solution steps

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

Expand

(x^{4} + 5)^{3} x^{3} : x^{15} + 15 x^{11} + 75 x^{7} + 125 x^{3}

= \int x^{15} + 15 x^{11} + 75 x^{7} + 125 x^{3} 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 x^{15} d x + \int 15 x^{11} d x + \int 75 x^{7} d x + \int 125 x^{3} d x

\int x^{15} d x = \frac{x ^{16}}{16}

\int 15 x^{11} d x = \frac{5 x ^{12}}{4}

\int 75 x^{7} d x = \frac{75 x ^{8}}{8}

\int 125 x^{3} d x = \frac{125 x ^{4}}{4}

= \frac{x ^{16}}{16} + \frac{5 x ^{12}}{4} + \frac{75 x ^{8}}{8} + \frac{125 x ^{4}}{4}

Add a constant to the solution

= \frac{x ^{16}}{16} + \frac{5 x ^{12}}{4} + \frac{75 x ^{8}}{8} + \frac{125 x ^{4}}{4} + C

\frac{d}{d x} (cos^{2} (x) sin^{2} (x))

\int 9 r cos (r) d r

y = x^{4 x}

\frac{d y}{d x} = (y^{2} - 4)

x \to - 2 lim (x + 1)

What is the integral of (x^4+5)^3x^3 ?
The integral of (x^4+5)^3x^3 is (x^{16}}{16}+\frac{5x^{12})/4+(75x^8)/8+(125x^4)/4+C