Question

How do you integrate $\int {\dfrac{{x{e^x}}}{{{{\left( {x + 1} \right)}^2}}}} dx$ using integration by parts?

Answer 1

We have to find $\int {\dfrac{{x{e^x}}}{{{{\left( {x + 1} \right)}^2}}}} dx$ using integration by parts. As we know, integration by parts or partial integration is a process that finds the integral of a product of function in terms of the integral of the product of their derivative and antiderivative. We will first write the $x$ in the numerator as $\left( {x + 1 - 1} \right)$, then we will divide the numerator with the denominator. We will apply integration by parts on $\int {\dfrac{{{e^x}}}{{\left( {x + 1} \right)}}} dx$ and then we will cancel the common terms to find the result.

Complete answer:
As we know, integration by parts or partial integration is a process that finds the integral of a product of function in terms of the integral of the product of their derivative and antiderivative.
The rule can be thought of as an integral version of the product rule of differentiation.
The integration by parts formula states:
$\int {u(x)v'(x)dx} = u(x)v(x) - \int {u'(x)v(x)dx}$
Let, $I = \int {\dfrac{{x{e^x}}}{{{{\left( {x + 1} \right)}^2}}}} dx$
We can write the $x$ in the numerator as $\left( {x + 1 - 1} \right)$.
So, we get
$\Rightarrow I = \int {\dfrac{{\left( {x + 1 - 1} \right){e^x}}}{{{{\left( {x + 1} \right)}^2}}}} dx$
On dividing, we get
$\Rightarrow I = \int {\left[ {\dfrac{{x + 1}}{{{{\left( {x + 1} \right)}^2}}} - \dfrac{1}{{{{\left( {x + 1} \right)}^2}}}} \right]} {e^x}dx$
Cancelling the common terms from the numerator and the denominator,
$\Rightarrow I = \int {\left[ {\dfrac{1}{{\left( {x + 1} \right)}} - \dfrac{1}{{{{\left( {x + 1} \right)}^2}}}} \right]{e^x}} dx$
On multiplying the terms, we get
$\Rightarrow I = \int {\dfrac{{{e^x}}}{{\left( {x + 1} \right)}}} dx - \int {\dfrac{{{e^x}}}{{{{\left( {x + 1} \right)}^2}}}} dx$
Applying integration by parts in first integral i.e., $\int {\dfrac{{{e^x}}}{{\left( {x + 1} \right)}}} dx$, we get
$\Rightarrow I = \left( {\dfrac{1}{{\left( {x + 1} \right)}} \times \int {{e^x}} dx} \right) - \int {\left( {\dfrac{d}{{dx}}\left( {\dfrac{1}{{\left( {x + 1} \right)}}} \right) \times \int {{e^x}dx} } \right)dx} - \int {\dfrac{{{e^x}}}{{{{\left( {x + 1} \right)}^2}}}} dx + C$
As we know that $\int {{e^x}dx = } {e^x}$ and $\dfrac{d}{{dx}}\left( {\dfrac{1}{{\left( {x + 1} \right)}}} \right) = \dfrac{{ - 1}}{{{{\left( {x + 1} \right)}^2}}}$, putting this we get
$\Rightarrow I = \left( {\dfrac{1}{{\left( {x + 1} \right)}} \times {e^x}} \right) - \int {\left( {\left( { - \dfrac{1}{{{{\left( {x + 1} \right)}^2}}}} \right) \times {e^x}} \right)dx} - \int {\dfrac{{{e^x}}}{{{{\left( {x + 1} \right)}^2}}}} dx + C$
On simplifying, we get
$\Rightarrow I = \dfrac{{{e^x}}}{{\left( {x + 1} \right)}} + \int {\dfrac{{{e^x}}}{{{{\left( {x + 1} \right)}^2}}}} dx - \int {\dfrac{{{e^x}}}{{{{\left( {x + 1} \right)}^2}}}} dx + C$
As we can see that the second term and the third term are the same with opposite signs. Therefore, we can cancel them. So, we get
$\Rightarrow I = \dfrac{{{e^x}}}{{\left( {x + 1} \right)}} + C$
Therefore, we get $\int {\dfrac{{x{e^x}}}{{{{\left( {x + 1} \right)}^2}}}} dx$ using integration by parts as $\dfrac{{{e^x}}}{{\left( {x + 1} \right)}} + C$.

Note:
Here, we get a constant of integration $'C'$ because this is a case of indefinite integration. In the case of definite integration where lower and upper limits are given, integration by parts formula can be written as: $\int\limits_a^b {u(x)v'(x)dx} = \left[ {u(x)v(x)} \right]_a^b - \int\limits_a^b {u'(x)v(x)dx}$ i.e., $\int\limits_a^b {u(x)v'(x)dx} = u(b)v(b) - u(a)v(a) - \int\limits_a^b {u'(x)v(x)dx}$ where $a$ is the lower limit and $b$ is the upper limit.