What is the integral of x^4(x+1)^4(2x+1) ?

The integral of x^4(x+1)^4(2x+1) is (x^{10})/5+x^9+2x^8+2x^7+x^6+(x^5)/5+C

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integral of x^4(x+1)^4(2x+1)

\int x^{4} (x + 1)^{4} (2 x + 1) d x

\frac{x ^{10}}{5} + x^{9} + 2 x^{8} + 2 x^{7} + x^{6} + \frac{x ^{5}}{5} + C

Solution steps

\int x^{4} (x + 1)^{4} (2 x + 1) d x

Expand

x^{4} (x + 1)^{4} (2 x + 1) : 2 x^{9} + 9 x^{8} + 16 x^{7} + 14 x^{6} + 6 x^{5} + x^{4}

= \int 2 x^{9} + 9 x^{8} + 16 x^{7} + 14 x^{6} + 6 x^{5} + x^{4} 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 2 x^{9} d x + \int 9 x^{8} d x + \int 16 x^{7} d x + \int 14 x^{6} d x + \int 6 x^{5} d x + \int x^{4} d x

\int 2 x^{9} d x = \frac{x ^{10}}{5}

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

\int 16 x^{7} d x = 2 x^{8}

\int 14 x^{6} d x = 2 x^{7}

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

\int x^{4} d x = \frac{x ^{5}}{5}

= \frac{x ^{10}}{5} + x^{9} + 2 x^{8} + 2 x^{7} + x^{6} + \frac{x ^{5}}{5}

Add a constant to the solution

= \frac{x ^{10}}{5} + x^{9} + 2 x^{8} + 2 x^{7} + x^{6} + \frac{x ^{5}}{5} + C

y^{^{'}} = x - y

(2.9) (5.8)

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

\int sin^{3} (x) cos^{4} (x) d x

\frac{d}{d x} (\frac{x sin ( x )}{x ^{2} - 1})

What is the integral of x^4(x+1)^4(2x+1) ?
The integral of x^4(x+1)^4(2x+1) is (x^{10})/5+x^9+2x^8+2x^7+x^6+(x^5)/5+C