Question

Find the correct answer.
$\int {\dfrac{{x{e^x}}}{{{{\left( {1 + x} \right)}^2}}}} $ is equal to which of the following?
1) $\dfrac{{{e^x}}}{{\left( {1 + x} \right)}} + c$
2) $\dfrac{{{e^x}}}{{{{\left( {1 + x} \right)}^2}}} + c$
3) ${e^x}\left( {1 + x} \right) + c$
4) $\dfrac{{ - {e^x}}}{{{{\left( {1 + x} \right)}^2}}} + c$

Answer 1

Hint: Above problem is based on the concept of integration by parts, which is given as stated below:
$\int {v\dfrac{{du}}{{dx}}} dx = uv - \int {u\dfrac{{dv}}{{dx}}} dx$
Integration of first function and then again integration of the same function with derivation of the second function.
Using the above concept we will solve the given problem.

Complete step-by-step answer:
Let's discuss the concept of integration by parts in more details and then we will proceed for the calculation part.
When we are required to find the integration of a function which contains two functions simultaneously then, we use integration by parts concept.
We use it to examine the function and find out which of the two functions is easily integrable and which function is derivable easily. After examining we apply the formula of integration by parts which is stated below:
$\int {v\dfrac{{du}}{{dx}}} dx = uv - \int {u\dfrac{{dv}}{{dx}}} dx$
Now we will calculate the given function using integration by parts.
$ \Rightarrow \int {\dfrac{{(1 - 1 + x){e^x}}}{{{{(1 + x)}^2}}}} $ (We have added and subtracted one in the numerator to simplify the given function)$ \Rightarrow \int {\dfrac{{{e^x}}}{{(1 + x)}}} - \int {\dfrac{{{e^x}}}{{{{(1 + x)}^2}}}} $ (We have multiplied the exponential term inside the bracket and obtained two different functions)
Now apply integration by parts, in first integral function;
$ \Rightarrow \int {\dfrac{{{e^x}}}{{(1 + x)}}dx} - \int {{e^x} \times \dfrac{d}{{dx}}\dfrac{1}{{(1 + x)}}} dx - \int {\dfrac{{{e^x}}}{{{{(1 + x)}^2}}}} dx$ (We have integrated the first function which is exponential and then we have differentiated the second function in the second term and we have let remain the second function as it is)
$ \Rightarrow \int {\dfrac{{{e^x}}}{{(1 + x)}}dx} - \int {{e^x}\dfrac{{ - 1}}{{{{(1 + x)}^2}}}} dx - \int {\dfrac{{{e^x}}}{{{{(1 + x)}^2}}}} dx$
$ \Rightarrow \dfrac{{{e^x}}}{{(1 + x)}}$ (We have cancelled the common term)
$ \Rightarrow \dfrac{{{e^x}}}{{(1 + x)}}$+c

Option 1 is correct.

Note: Integration by parts has many other applications in problem solving such as in deriving the Euler-Lagrange equation, for determining the boundary conditions in Sturm-Liouville theory, it is used in operator theory, decay of Fourier transform, Fourier transform of derivative, used in harmonics analysis, to find the gamma function identity etc.