What is the integral from 0 to 1 of (x^2)/((x+1)^3) ?

The integral from 0 to 1 of (x^2)/((x+1)^3) is (16ln(2)+30)/(16)-5/2

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integral from 0 to 1 of (x^2)/((x+1)^3)

\int_{0}^{1} \frac{x ^{2}}{( x + 1 ) ^{3}} d x

\frac{16 ln ( 2 ) + 30}{16} - \frac{5}{2}

Decimal Notation

0.06814 \dots

Solution steps

\int_{0}^{1} \frac{x ^{2}}{( x + 1 ) ^{3}} d x

Apply Integration By Parts

: [\frac{x ^{3}}{( x + 1 ) ^{3}} - \int \frac{x ^{2} ( - x + 2 )}{( x + 1 ) ^{4}} d x]_{0}^{1}

= [\frac{x ^{3}}{( x + 1 ) ^{3}} - \int \frac{x ^{2} ( - x + 2 )}{( x + 1 ) ^{4}} d x]_{0}^{1}

\int \frac{x ^{2} ( - x + 2 )}{( x + 1 ) ^{4}} d x = - ln ∣ x + 1 ∣ - \frac{3}{x + 1} + \frac{3}{2 ( x + 1 ) ^{2}} - \frac{1}{3 ( x + 1 ) ^{3}} - \frac{2 x ^{2}}{3 ( x + 1 ) ^{3}} + \frac{2 ( - 2 x - 1 )}{3 ( x + 1 ) ^{2}}

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

Simplify

[\frac{x ^{3}}{( x + 1 ) ^{3}} - (- ln ∣ x + 1 ∣ - \frac{3}{x + 1} + \frac{3}{2 ( x + 1 ) ^{2}} - \frac{1}{3 ( x + 1 ) ^{3}} - \frac{2 x ^{2}}{3 ( x + 1 ) ^{3}} + \frac{2 ( - 2 x - 1 )}{3 ( x + 1 ) ^{2}})]_{0}^{1} : [\frac{2 x ^{3} + 10 x ^{2} + 13 x + 2 ln ∣ x + 1 ∣ ( x + 1 ) ^{3} + 5}{2 ( x + 1 ) ^{3}}]_{0}^{1}

= [\frac{2 x ^{3} + 10 x ^{2} + 13 x + 2 ln ∣ x + 1 ∣ ( x + 1 ) ^{3} + 5}{2 ( x + 1 ) ^{3}}]_{0}^{1}

Compute the boundaries:

\frac{16 ln ( 2 ) + 30}{16} - \frac{5}{2}

= \frac{16 ln ( 2 ) + 30}{16} - \frac{5}{2}

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What is the integral from 0 to 1 of (x^2)/((x+1)^3) ?
The integral from 0 to 1 of (x^2)/((x+1)^3) is (16ln(2)+30)/(16)-5/2