What is the integral of (x^2)/((1+x)^8) ?

The integral of (x^2)/((1+x)^8) is -1/(5(1+x)^5)+1/(3(1+x)^6)-1/(7(1+x)^7)+C

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integral of (x^2)/((1+x)^8)

\int \frac{x ^{2}}{( 1 + x ) ^{8}} d x

- \frac{1}{5 ( 1 + x ) ^{5}} + \frac{1}{3 ( 1 + x ) ^{6}} - \frac{1}{7 ( 1 + x ) ^{7}} + C

Solution steps

\int \frac{x ^{2}}{( 1 + x ) ^{8}} d x

Apply u-substitution

: \int \frac{( u - 1 ) ^{2}}{u ^{8}} d u

= \int \frac{( u - 1 ) ^{2}}{u ^{8}} d u

Expand

\frac{( u - 1 ) ^{2}}{u ^{8}} : \frac{1}{u ^{6}} - \frac{2}{u ^{7}} + \frac{1}{u ^{8}}

= \int \frac{1}{u ^{6}} - \frac{2}{u ^{7}} + \frac{1}{u ^{8}} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{1}{u ^{6}} d u - \int \frac{2}{u ^{7}} d u + \int \frac{1}{u ^{8}} d u

\int \frac{1}{u ^{6}} d u = - \frac{1}{5 u ^{5}}

\int \frac{2}{u ^{7}} d u = - \frac{1}{3 u ^{6}}

\int \frac{1}{u ^{8}} d u = - \frac{1}{7 u ^{7}}

= - \frac{1}{5 u ^{5}} - (- \frac{1}{3 u ^{6}}) - \frac{1}{7 u ^{7}}

Substitute back

u = 1 + x

= - \frac{1}{5 ( 1 + x ) ^{5}} - (- \frac{1}{3 ( 1 + x ) ^{6}}) - \frac{1}{7 ( 1 + x ) ^{7}}

Simplify

= - \frac{1}{5 ( 1 + x ) ^{5}} + \frac{1}{3 ( 1 + x ) ^{6}} - \frac{1}{7 ( 1 + x ) ^{7}}

Add a constant to the solution

= - \frac{1}{5 ( 1 + x ) ^{5}} + \frac{1}{3 ( 1 + x ) ^{6}} - \frac{1}{7 ( 1 + x ) ^{7}} + C

\int x^{3} x^{2} + 13 d x

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What is the integral of (x^2)/((1+x)^8) ?
The integral of (x^2)/((1+x)^8) is -1/(5(1+x)^5)+1/(3(1+x)^6)-1/(7(1+x)^7)+C