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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{1}{24}

Decimal Notation

0.04166 \dots

Solution steps

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

Apply Integration By Parts

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

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

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

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

Simplify

[- \frac{x ^{2}}{3 ( 1 + x ) ^{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}

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

\int_{1}^{x} e^{- x^{2}} d x

\int_{0}^{3} x (x - 1) (x - 3) d x

\int_{1}^{x} \frac{3}{x ^{4}} d x

\int_{- 8}^{- 3} - x^{2} - 3 x d x

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