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

The integral from 1 to 2 of x/((x+1)^2) is ln(3)-ln(2)-1/6

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

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

ln (3) - ln (2) - \frac{1}{6}

Decimal Notation

0.23879 \dots

Solution steps

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

Apply u-substitution

: \int_{2}^{3} \frac{u - 1}{u ^{2}} d u

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

Expand

\frac{u - 1}{u ^{2}} : \frac{1}{u} - \frac{1}{u ^{2}}

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

\int_{2}^{3} \frac{1}{u} d u = ln (3) - ln (2)

\int_{2}^{3} \frac{1}{u ^{2}} d u = \frac{1}{6}

= ln (3) - ln (2) - \frac{1}{6}

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

y^{^{'}} = e^{4 x} - 3 x

\frac{d}{d x} (2 cot (\frac{x}{2}))

y^{^{'}} = x^{2} - y

36 y^{^{''}} - y = x e^{\frac{x}{6}}, y (0) = 1, y^{^{'}} (0) = 0

What is the integral from 1 to 2 of x/((x+1)^2) ?
The integral from 1 to 2 of x/((x+1)^2) is ln(3)-ln(2)-1/6