What is the integral from 0 to 1 of (x^2)/((x+1)^4) ?

The integral from 0 to 1 of (x^2)/((x+1)^4) is 1/24

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integral from 0 to 1 of (x^2)/((x+1)^4)

\int_{0}^{1} \frac{x ^{2}}{( x + 1 ) ^{4}} d x

\frac{1}{24}

Decimal

0.04166 \dots

Solution steps

\int_{0}^{1} \frac{x ^{2}}{( x + 1 ) ^{4}} d x

Apply Integration By Parts

: [- \frac{x ^{2}}{3 ( x + 1 ) ^{3}} - \int - \frac{2 x}{3 ( x + 1 ) ^{3}} d x]_{0}^{1}

= [- \frac{x ^{2}}{3 ( x + 1 ) ^{3}} - \int - \frac{2 x}{3 ( x + 1 ) ^{3}} d x]_{0}^{1}

\int - \frac{2 x}{3 ( x + 1 ) ^{3}} d x = - \frac{- 2 x - 1}{3 ( x + 1 ) ^{2}}

= [- \frac{x ^{2}}{3 ( x + 1 ) ^{3}} - (- \frac{- 2 x - 1}{3 ( x + 1 ) ^{2}})]_{0}^{1}

Simplify

[- \frac{x ^{2}}{3 ( x + 1 ) ^{3}} - (- \frac{- 2 x - 1}{3 ( x + 1 ) ^{2}})]_{0}^{1} : [\frac{- 3 x ^{2} - 3 x - 1}{3 ( x + 1 ) ^{3}}]_{0}^{1}

= [\frac{- 3 x ^{2} - 3 x - 1}{3 ( x + 1 ) ^{3}}]_{0}^{1}

Compute the boundaries:

\frac{1}{24}

= \frac{1}{24}

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What is the integral from 0 to 1 of (x^2)/((x+1)^4) ?
The integral from 0 to 1 of (x^2)/((x+1)^4) is 1/24