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

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

Popular >

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

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

3

Solution steps

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

Apply Integration By Parts

: [\frac{4 x}{( x + 1 ) ^{2}} - \int - \frac{8 x}{( x + 1 ) ^{3}} d x]_{0}^{3}

= [\frac{4 x}{( x + 1 ) ^{2}} - \int - \frac{8 x}{( x + 1 ) ^{3}} d x]_{0}^{3}

\int - \frac{8 x}{( x + 1 ) ^{3}} d x = - 8 (- \frac{1}{x + 1} + \frac{1}{2 ( x + 1 ) ^{2}})

= [\frac{4 x}{( x + 1 ) ^{2}} - (- 8 (- \frac{1}{x + 1} + \frac{1}{2 ( x + 1 ) ^{2}}))]_{0}^{3}

Simplify

= [\frac{4 x}{( x + 1 ) ^{2}} + 8 (- \frac{1}{x + 1} + \frac{1}{2 ( x + 1 ) ^{2}})]_{0}^{3}

Compute the boundaries:

3

= 3

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

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

\int_{1}^{5} ln (3 x) - ln (x) d x

\int_{2}^{3} (x^{2} - 2 x) d x

\int_{- 2}^{4} \frac{x}{2} d x

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