Question
Question: Find: \[\int{\dfrac{\left( 2x-5 \right){{e}^{2x}}}{{{\left( 2x-3 \right)}^{3}}}}dx\]...
Find: ∫(2x−3)3(2x−5)e2xdx
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, we can integrate I1 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=∫(2x−3)3(2x−5)e2xdx
We can now split the numerator as,
⇒I=∫(2x−3)3[(2x−3)−2]e2xdx
We can now multiply the exponential term inside the bracket in the numerator and split the integral into two parts, we get
⇒I=∫(2x−3)3(2x−3)e2xdx−∫(2x−3)32e2xdx
We can now cancel the similar terms in the numerator and the denominator, we get
⇒I=∫(2x−3)2e2xdx−∫(2x−3)32e2xdx
We can consider the first term as,
⇒I1=∫(2x−3)2e2xdx
Now we can integrate I1 by u-v method, by keeping algebraic term as u and exponential term as dv as by the ILATE rule,
Let
u=(2x−3)21 , dv=e2xdx
Then,
dxdu=2(2x−3)−2, v=2e2x
We can now substitute the above values in the formula uv−∫vdu, we get
⇒(2x−3)212e2x−∫[dxd(2x−3)−22e2x]dx
We can now integrate the above step, we get
⇒(2x−3)212e2x−∫(−2)(2x−3)−322e2xdx
We can now write the above terms as,
⇒I1=(2x−3)212e2x+∫(2x−3)32e2xdx+C
We can now write in simple terms as,
⇒I1=(2x−3)212e2x+I2+C……. (1)
We know that, I=I1−I2.
We can now substitute (10) in the above step, we get
⇒I1=(2x−3)212e2x+I2−I2+C
We can now simplify the above step, we get
⇒I1=2(2x−3)2e2x+C
Therefore, the integrated form is I1=2(2x−3)2e2x+C.
Note: Students should always remember the formula of the u-v method is uv−∫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.