What is the integral from 0 to 9 of (10x)/((1+x)^2) ?

The integral from 0 to 9 of (10x)/((1+x)^2) is 10ln(10)-9

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integral from 0 to 9 of (10x)/((1+x)^2)

\int_{0}^{9} \frac{10 x}{( 1 + x ) ^{2}} d x

10 ln (10) - 9

Decimal

14.02585 \dots

Solution steps

\int_{0}^{9} \frac{10 x}{( 1 + x ) ^{2}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 10 \cdot \int_{0}^{9} \frac{x}{( 1 + x ) ^{2}} d x

Apply u-substitution

: \int_{1}^{10} \frac{u - 1}{u ^{2}} d u

= 10 \cdot \int_{1}^{10} \frac{u - 1}{u ^{2}} d u

Expand

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

= 10 \cdot \int_{1}^{10} \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

= 10 (\int_{1}^{10} \frac{1}{u} d u - \int_{1}^{10} \frac{1}{u ^{2}} d u)

\int_{1}^{10} \frac{1}{u} d u = ln (10)

\int_{1}^{10} \frac{1}{u ^{2}} d u = \frac{9}{10}

= 10 (ln (10) - \frac{9}{10})

Simplify

10 (ln (10) - \frac{9}{10}) : 10 ln (10) - 9

= 10 ln (10) - 9

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What is the integral from 0 to 9 of (10x)/((1+x)^2) ?
The integral from 0 to 9 of (10x)/((1+x)^2) is 10ln(10)-9