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

The integral from 0 to 1 of x/((x+1)^4) is 1/12

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

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

\frac{1}{12}

Decimal

0.08333 \dots

Solution steps

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

Apply u-substitution

= \int_{1}^{2} \frac{u - 1}{u ^{4}} d u

Expand

\frac{u - 1}{u ^{4}} : \frac{1}{u ^{3}} - \frac{1}{u ^{4}}

= \int_{1}^{2} \frac{1}{u ^{3}} - \frac{1}{u ^{4}} 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_{1}^{2} \frac{1}{u ^{3}} d u - \int_{1}^{2} \frac{1}{u ^{4}} d u

\int_{1}^{2} \frac{1}{u ^{3}} d u = \frac{3}{8}

\int_{1}^{2} \frac{1}{u ^{4}} d u = \frac{7}{24}

= \frac{3}{8} - \frac{7}{24}

Simplify

\frac{3}{8} - \frac{7}{24} : \frac{1}{12}

= \frac{1}{12}

\int_{0}^{4} sin (\frac{nπ x}{4}) d x

\int_{0}^{3} 12 d x

\int_{1}^{4} 119 x^{\frac{5}{2}} d x

\int_{2}^{3} e^{x^{3}} x^{2} d x

\int_{7}^{9} \frac{14}{( x - 8 ) ^{2}} d x

What is the integral from 0 to 1 of x/((x+1)^4) ?
The integral from 0 to 1 of x/((x+1)^4) is 1/12