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Found in: Page 363

### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

# $${\bf{1 - 40}}$$ - Evaluate the integral.$$\int_1^2 {\frac{x}{{{{(x + 1)}^2}}}} dx$$

$$\int_1^2 {\frac{x}{{{{(x + 1)}^2}}}} dx = \ln 3 - \ln 2 - \frac{1}{6}$$

See the step by step solution

## Step 1: Definition

Integration is a method of adding or summing up the parts to find the whole. It is a reverse process of differentiation.

## Step 2: Substitute

Consider the definite integral$$\int_1^2 {\frac{x}{{{{(x + 1)}^2}}}} dx$$

Let$$t = x + 1$$.

$$\Rightarrow dt = dx$$

Limit changes to:

For$$x = 1$$

\begin{aligned}{c}t &= 1 + 1\\ &= 2\end{aligned}

And for$$x = 2$$

\begin{aligned}{c}t &= 2 + 1\\ &= 3\end{aligned}

## Step 3: Evaluate integral

Replacing values we have:

\begin{aligned}{c}\int_1^2 {\frac{x}{{{{(x + 1)}^2}}}} dx &= \int_2^3 {\left( {\frac{{u - 1}}{{{u^2}}}} \right)} du\\ &= \int_2^3 {\left( {\frac{1}{u} - \frac{1}{{{u^2}}}} \right)} du\\ &= \left( {\ln |u| + \frac{1}{u}} \right)_2^3\\ &= \left( {\ln 3 + \frac{1}{3}} \right) - \left( {\ln 2 + \frac{1}{2}} \right)\\ &= \ln 3 - \ln 2 - \frac{1}{6}\end{aligned}

Therefore, the value of the given integral is$$\ln 3 - \ln 2 - \frac{1}{6}$$