Question

Evaluate the integral:\(\int\limits_0^1 {\frac{{2x + 3}}{{{{(x + 1)}^2}}}} \)

Short answer

Hence, the solution is \(\frac{1}{2}\).

Step-by-step solution

Step-1: Using the partial function,

\(\frac{{2x + 3}}{{{{(x + 1)}^2}}}\) Can be written as \(\frac{A}{{(x + 1)}} + \frac{B}{{{{(x + 1)}^2}}}\)

Step-2:  Solving

\(\begin{aligned}{l}\frac{{2x + 3}}{{{{(x + 1)}^2}}} = \frac{A}{{x + 1}} + \frac{B}{{{{(x + 1)}^2}}}\\\frac{{2x + 3}}{{{{(x + 1)}^2}}} = \frac{{A(x + 1) + B}}{{{{(x + 1)}^2}}}\\2x + 3 = Ax + A + B\\A = 2\\A + B = 3 \Rightarrow B = 3 - A = 3 - 2 = 1\end{aligned}\)

Step-3: Applying the formula

\(\begin{aligned}{l}\int {\frac{2}{{x + 1}}dx} + \int {\frac{1}{{{{(x + 1)}^2}}}dx} \\\int {\frac{2}{{x + 1}}dx} \Rightarrow x + 1 = w\\1 = \frac{{dw}}{{dx}}\\ \Rightarrow dx = dw\\\int {\frac{{2dw}}{w} = 2\log \left| w \right| = 2\log \left| {x + 1} \right|} \end{aligned}\)

Step-4: Solving the problem

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

Let\((x + 1) = t\)

\(\begin{aligned}{l}1 = \frac{{dt}}{{dx}} \Rightarrow dx = dt\\\int {\frac{{dt}}{{{t^2}}} = \int {{t^{ - 2}}dt = \frac{{{t^{ - 2 + 1}}}}{{ - 2 + 1}}} } \\ = \frac{{{t^{ - 1}}}}{{ - 1}} = \frac{{ - 1}}{t} = \frac{{ - 1}}{{x + 1}}\end{aligned}\)

Step-5:  Putting the limits

\(\begin{aligned}{l}\left( {\frac{{ - 1}}{{x + 1}}} \right)_0^1 = \frac{{ - 1}}{{1 + 1}} - \left( {\frac{{ - 1}}{{0 + 1}}} \right)\\ = \frac{{ - 1}}{2} - \left( {\frac{{ - 1}}{1}} \right) = \frac{{ - 1}}{2} + 1\\ = 1 - \frac{1}{2} = \frac{1}{2}\end{aligned}\)

Hence, the solution is \(\frac{1}{2}\).