Question

Find: \[\int{\dfrac{\left( 2x-5 \right){{e}^{2x}}}{{{\left( 2x-3 \right)}^{3}}}}dx\]

Answer 1

Hint: In this problem, we have to integrate the given integral. We can first split the given integral into two parts, we can then take the first term as \[{{I}_{1}}\], we can integrate \[{{I}_{1}}\] by u-v method, by keeping algebraic term as u and exponential term as dv as by the ILATE rule. We can then simplify the terms to get an integrated form.

Complete step by step answer:
Let us assume the given equation
\[\Rightarrow I=\int{\dfrac{\left( 2x-5 \right){{e}^{2x}}}{{{\left( 2x-3 \right)}^{3}}}}dx\]
We can now split the numerator as,
\[\Rightarrow I=\int{\dfrac{\left[ \left( 2x-3 \right)-2 \right]{{e}^{2x}}}{{{\left( 2x-3 \right)}^{3}}}}dx\]
We can now multiply the exponential term inside the bracket in the numerator and split the integral into two parts, we get
\[\Rightarrow I=\int{\dfrac{\left( 2x-3 \right){{e}^{2x}}}{{{\left( 2x-3 \right)}^{3}}}}dx-\int{\dfrac{2{{e}^{2x}}}{{{\left( 2x-3 \right)}^{3}}}}dx\]
We can now cancel the similar terms in the numerator and the denominator, we get
\[\Rightarrow I=\int{\dfrac{{{e}^{2x}}}{{{\left( 2x-3 \right)}^{2}}}}dx-\int{\dfrac{2{{e}^{2x}}}{{{\left( 2x-3 \right)}^{3}}}}dx\]
We can consider the first term as,
\[\Rightarrow {{I}_{1}}=\int{\dfrac{{{e}^{2x}}}{{{\left( 2x-3 \right)}^{2}}}dx}\]
Now we can integrate \[{{I}_{1}}\] by u-v method, by keeping algebraic term as u and exponential term as dv as by the ILATE rule,
Let
\[u=\dfrac{1}{{{\left( 2x-3 \right)}^{2}}}\] , \[dv={{e}^{2x}}dx\]
Then,
\[\dfrac{du}{dx}=\dfrac{{{\left( 2x-3 \right)}^{-2}}}{2}\], \[v=\dfrac{{{e}^{2x}}}{2}\]
We can now substitute the above values in the formula \[uv-\int{vdu}\], we get
 \[\Rightarrow \dfrac{1}{{{\left( 2x-3 \right)}^{2}}}\dfrac{{{e}^{2x}}}{2}-\int{\left[ \dfrac{d}{dx}{{\left( 2x-3 \right)}^{-2}}\dfrac{{{e}^{2x}}}{2} \right]}dx\]
We can now integrate the above step, we get
\[\Rightarrow \dfrac{1}{{{\left( 2x-3 \right)}^{2}}}\dfrac{{{e}^{2x}}}{2}-\int{\left( -2 \right)}{{\left( 2x-3 \right)}^{-3}}2\dfrac{{{e}^{2x}}}{2}dx\]
We can now write the above terms as,
\[\Rightarrow {{I}_{1}}=\dfrac{1}{{{\left( 2x-3 \right)}^{2}}}\dfrac{{{e}^{2x}}}{2}+\int{\dfrac{2e^{2x}}{{{\left( 2x-3 \right)}^{3}}}}dx+C\]
We can now write in simple terms as,
\[\Rightarrow {{I}_{1}}=\dfrac{1}{{{\left( 2x-3 \right)}^{2}}}\dfrac{{{e}^{2x}}}{2}+{{I}_{2}}+C\]……. (1)
 We know that, \[I={{I}_{1}}-{{I}_{2}}\].
We can now substitute (10) in the above step, we get
\[\Rightarrow {{I}_{1}}=\dfrac{1}{{{\left( 2x-3 \right)}^{2}}}\dfrac{{{e}^{2x}}}{2}+{{I}_{2}}-{{I}_{2}}+C\]
We can now simplify the above step, we get
\[\Rightarrow {{I}_{1}}=\dfrac{{{e}^{2x}}}{2{{\left( 2x-3 \right)}^{2}}}+C\]

Therefore, the integrated form is \[{{I}_{1}}=\dfrac{{{e}^{2x}}}{2{{\left( 2x-3 \right)}^{2}}}+C\].

Note: Students should always remember the formula of the u-v method is \[uv-\int{vdu}\], where we should use the ILATE rule to choose u and dv, we can then differentiate u to get du and integrate dv to get v and substitute it in the formula. We should integrate the resulting step to get the final answer.