Question

\({\bf{1 - 40}}\)- Evaluate the integral.

\(\int_1^2 {\frac{x}{{{{(x + 1)}^2}}}} dx\)

Short answer

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

Step-by-step solution

Definition

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

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\(x = 2\)

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

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