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.