Question

Find the value of \(\smallint \frac{{x \cdot {e^x}}}{{{{\left( {1 + x} \right)}^2}}}dx\)
1. \(\frac{{{e^x}}}{{1 + x}^2} + C\)
2. \(\frac{{{e^x}}}{{1 + x}} + C\)
3. \(- \ \frac{{{e^x}}}{{1 + x}} + C\)
4. \(- \ \frac{{{e^x}}}{{1 + x}^2} + C\)
5. None of these

Answer 1

Correct Answer - Option 2 : \(\frac{{{e^x}}}{{1 + x}} + C\)

Concept:

• \(\smallint {e^x} \cdot \left[ {f\left( x \right) + f'\left( x \right)} \right]\;dx = {e^x} \cdot f\left( x \right) + C\) where C is a constant
• \(\frac{{d\left( {{x^n}} \right)}}{{dx}} = n \cdot {x^{n - 1}}\)

Calculation:

Here we have to find the value of \(\smallint \frac{{x \cdot {e^x}}}{{{{\left( {1 + x} \right)}^2}}}dx\)

Now the given integral can be re-written as:

\(\Rightarrow \smallint \frac{{x \cdot {e^x}}}{{{{\left( {1 + x} \right)}^2}}}dx = \;\smallint {e^x} \cdot \left\{ {\frac{{\left( {1 + x} \right) - 1}}{{{{\left( {1 + x} \right)}^2}}}} \right\}dx\)

\(\Rightarrow \smallint {e^x} \cdot \left\{ {\frac{{\left( {1 + x} \right) - 1}}{{{{\left( {1 + x} \right)}^2}}}} \right\}dx = \;\smallint {e^x} \cdot \left\{ {\frac{1}{{1 + x}} - \frac{1}{{{{\left( {1 + x} \right)}^2}}}} \right\} dx\)

Now by comparing the above equation with \(\smallint {e^x} \cdot \left[ {f\left( x \right) + f'\left( x \right)} \right]\;dx\) we get

\(f\left( x \right) = \;\frac{1}{{1 + x}}\;\;and\;f'\left( x \right) = \;\frac{1}{{{{\left( {1 + x} \right)}^2}}}\)

As we know that, \(\smallint {e^x} \cdot \left[ {f\left( x \right) + f'\left( x \right)} \right]\;dx = {e^x} \cdot f\left( x \right) + C\) where C is a constant

\(\Rightarrow \smallint \frac{{x \cdot {e^x}}}{{{{\left( {1 + x} \right)}^2}}}dx = \frac{{{e^x}}}{{1 + x}} + C\) where C is a constant