What is the integral from 0 to 1 of x/((x+31)^2) ?

The integral from 0 to 1 of x/((x+31)^2) is 5ln(2)-ln(31)-1/32

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integral from 0 to 1 of x/((x+31)^2)

\int_{0}^{1} \frac{x}{( x + 31 ) ^{2}} d x

5 ln (2) - ln (31) - \frac{1}{32}

Decimal

0.00049 \dots

Solution steps

\int_{0}^{1} \frac{x}{( x + 31 ) ^{2}} d x

Apply u-substitution

: \int_{31}^{32} \frac{u - 31}{u ^{2}} d u

= \int_{31}^{32} \frac{u - 31}{u ^{2}} d u

Expand

\frac{u - 31}{u ^{2}} : \frac{1}{u} - \frac{31}{u ^{2}}

= \int_{31}^{32} \frac{1}{u} - \frac{31}{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_{31}^{32} \frac{1}{u} d u - \int_{31}^{32} \frac{31}{u ^{2}} d u

\int_{31}^{32} \frac{1}{u} d u = 5 ln (2) - ln (31)

\int_{31}^{32} \frac{31}{u ^{2}} d u = \frac{1}{32}

= 5 ln (2) - ln (31) - \frac{1}{32}

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What is the integral from 0 to 1 of x/((x+31)^2) ?
The integral from 0 to 1 of x/((x+31)^2) is 5ln(2)-ln(31)-1/32