Question

How do you integrate $\int {x{e^{ - 4x}}}$ by integration by parts method?

Answer 1

Hint : In order to determine the answer of above definite integral use the formula of integration by parts i.e. $\int {f(x)g'(x)dx = f(x)g(x) - \int {f'(x)g(x)dx} }$ or \int {uvdx = u\left\\{ {\int {vdx} } \right\\} - \int {\left\\{ {\dfrac{{du}}{{dx}}\int {vdx} } \right\\}dx} } and assume $f(x) = u = x$ and $g'(x) = v = {e^{ - 4x}}$ calculate $f'(x)$ and $g(x)$ and put them into the formula and integrate.

Complete step by step solution:
We are given a function $\int {x{e^{ - 4x}}}$ for which we have to find the integral using Integration by parts.
The formula for calculation of integration of parts is
$\int {f(x)g'(x)dx = f(x)g(x) - \int {f'(x)g(x)dx} }$
\int {uvdx = u\left\\{ {\int {vdx} } \right\\} - \int {\left\\{ {\dfrac{{du}}{{dx}}\int {vdx} } \right\\}dx} }
In our question Let assume
$f(x) = u = x$
And $g'(x) = v = {e^{ - 4x}}$
Let’s find the values one by one:
$u = x$
$v = {e^{ - 4x}} \\\ \int {v = \int {\left( {{e^{ - 4x}}} \right)} } = \dfrac{{\left( {{e^{ - 4x}}} \right)}}{{ - 4}} \;$
$\dfrac{{du}}{{dx}} = \dfrac{{dx}}{{dx}} = 1$
Now put the values in the respective formula and we get:

\Rightarrow \int {uvdx = u\left\\{ {\int {vdx} } \right\\} - \int {\left\\{ {\dfrac{{du}}{{dx}}\int {vdx} } \right\\}dx} } \\\ \Rightarrow \int {x{e^{ - 4x}}dx = x\left\\{ {\int {{e^{ - 4x}}dx} } \right\\} - \int {\left\\{ {\dfrac{{dx}}{{dx}}\int {{e^{ - 4x}}dx} } \right\\}dx} } \\\ \Rightarrow \int {x{e^{ - 4x}}dx = x\left\\{ {\dfrac{{{e^{ - 4x}}}}{{ - 4}}} \right\\} - \int {\left\\{ {1.\dfrac{{{e^{ - 4x}}}}{{ - 4}}} \right\\}dx} } \\\ \Rightarrow \int {x{e^{ - 4x}}dx = \dfrac{{x{e^{ - 4x}}}}{{ - 4}} - \int {\dfrac{{{e^{ - 4x}}}}{{ - 4}}dx} } \;

We can again put the same value in the respective place and we get:

\Rightarrow \int {x{e^{ - 4x}}dx = \dfrac{{x{e^{ - 4x}}}}{{ - 4}} - \int {\dfrac{{{e^{ - 4x}}}}{{ - 4}}dx} } \\\ \Rightarrow \int {x{e^{ - 4x}}dx = - \dfrac{1}{4}x{e^{ - 4x}} - \dfrac{1}{{ - 4}}\int {{e^{ - 4x}}dx} } \\\ \Rightarrow \int {x{e^{ - 4x}}dx = - \dfrac{1}{4}x{e^{ - 4x}} + \dfrac{1}{4}} \left\\{ {\dfrac{{{e^{ - 4x}}}}{{ - 4}}} \right\\} + C \\\ \Rightarrow \int {x{e^{ - 4x}}dx = - \dfrac{1}{4}x{e^{ - 4x}} - } \dfrac{1}{{16}}{e^{ - 4x}} + C \;

Therefore, integration of $\int {x{e^{ - 4x}}}$ by using integration by parts method is
$- \dfrac{1}{4}x{e^{ - 4x}} - \dfrac{1}{{16}}{e^{ - 4x}} + C$
So, the correct answer is “ $- \dfrac{1}{4}x{e^{ - 4x}} - \dfrac{1}{{16}}{e^{ - 4x}} + C$ ”.

Note : 1. Different types of methods of Integration:
Integration by Substitution
Integration by parts
Integration of rational algebraic function by using partial fraction
2. Integration by Substitution: The method of evaluating the integral by reducing it to standard form by a proper substitution is called integration by substitution.