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Question

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

Find: (2x5)e2x(2x3)3dx\int{\dfrac{\left( 2x-5 \right){{e}^{2x}}}{{{\left( 2x-3 \right)}^{3}}}}dx

Explanation

Solution

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 I1{{I}_{1}}, we can integrate I1{{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
I=(2x5)e2x(2x3)3dx\Rightarrow I=\int{\dfrac{\left( 2x-5 \right){{e}^{2x}}}{{{\left( 2x-3 \right)}^{3}}}}dx
We can now split the numerator as,
I=[(2x3)2]e2x(2x3)3dx\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
I=(2x3)e2x(2x3)3dx2e2x(2x3)3dx\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
I=e2x(2x3)2dx2e2x(2x3)3dx\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,
I1=e2x(2x3)2dx\Rightarrow {{I}_{1}}=\int{\dfrac{{{e}^{2x}}}{{{\left( 2x-3 \right)}^{2}}}dx}
Now we can integrate I1{{I}_{1}} by u-v method, by keeping algebraic term as u and exponential term as dv as by the ILATE rule,
Let
u=1(2x3)2u=\dfrac{1}{{{\left( 2x-3 \right)}^{2}}} , dv=e2xdxdv={{e}^{2x}}dx
Then,
dudx=(2x3)22\dfrac{du}{dx}=\dfrac{{{\left( 2x-3 \right)}^{-2}}}{2}, v=e2x2v=\dfrac{{{e}^{2x}}}{2}
We can now substitute the above values in the formula uvvduuv-\int{vdu}, we get
1(2x3)2e2x2[ddx(2x3)2e2x2]dx\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
1(2x3)2e2x2(2)(2x3)32e2x2dx\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,
I1=1(2x3)2e2x2+2e2x(2x3)3dx+C\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,
I1=1(2x3)2e2x2+I2+C\Rightarrow {{I}_{1}}=\dfrac{1}{{{\left( 2x-3 \right)}^{2}}}\dfrac{{{e}^{2x}}}{2}+{{I}_{2}}+C……. (1)
We know that, I=I1I2I={{I}_{1}}-{{I}_{2}}.
We can now substitute (10) in the above step, we get
I1=1(2x3)2e2x2+I2I2+C\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
I1=e2x2(2x3)2+C\Rightarrow {{I}_{1}}=\dfrac{{{e}^{2x}}}{2{{\left( 2x-3 \right)}^{2}}}+C

Therefore, the integrated form is I1=e2x2(2x3)2+C{{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 uvvduuv-\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.