MHTCET 2016 Math Solved Paper

(b) Consider, I=∫(

4ex−25

2ex−5

)dx
∫(

4ex

2ex−5

)dx - ∫(

25

2ex−5

)dx
= 4∫(

ex

2ex−5

)dx - 25 ∫(

e−x

2−5e−x

)dx
Let 2 ex - 5 = u and 2 - 5e−x = v
⇒ 2exdx = du and 5e−xdx = dv
⇒ exdx=

du

2

and e−xdx=

dv

5

So, I=4‌∫

du

2u

- 25‌∫

du

5v

= 2‌log‌u−5‌log‌v+c
=2‌log(2ex−5) - 5‌log(2−5e−x)+c
= - 3 log (2ex - 5) + 5x + c
Therefore, I = 5x - 3 log(2ex - 5) + c
As, I = Ax + B log(2ex - 5) + c
Hence, A = 5 and B = - 3