Maharashtra Board 12th Commerce Maths Solutions Chapter 5 Integration Ex 5.3

Balbharati Maharashtra State Board Std 12 Commerce Statistics Part 1 Digest Pdf Chapter 5 Integration Ex 5.3 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 5 Integration Ex 5.3

Evaluate the following:

Question 1.
\(\int \frac{3 e^{2 t}+5}{4 e^{2 t}-5} d t\)
Solution:
Let I = \(\int \frac{3 e^{2 t}+5}{4 e^{2 t}-5} d t\)
Put, Numerator = A(Denominator) + B[\(\frac{d}{d x}\)(Denominator)]
∴ 3e^2t + 5 = A(4e^2t – 5) + B[\(\frac{d}{d t}\)(4e^2t – 5)]
∴ 3e^2t + 5 = A(4e^2t – 5) + B[4e^2t × 2 – 0]
∴ 3e^2t + 5 = (4A + 8B) e^2t – 5A
Equating the coefficient of e^2t and constant on both sides, we get
4A + 8B = 3
and -5A = 5
∴ A = -1
∴ from (1), 4(-1) + 8B = 3
∴ 8B = 7
∴ B = \(\frac{7}{8}\)

Question 2.
\(\int \frac{20-12 e^{x}}{3 e^{x}-4} d x\)
Solution:
Let I = \(\int \frac{20-12 e^{x}}{3 e^{x}-4} d x\)
Put, Numerator = A (Denominator) + B[\(\frac{d}{d x}\)(Denominator)]
∴ 20 – 12e^x = A(3e^x – 4) + B[\(\frac{d}{d x}\)(3e^x – 4)]
∴ 20 – 12e^x = A(3e^x – 4) + B(3e^x – 0)
∴ 20 – 12e^x = (3A + 3B)e^x – 4A
Equating the coefficient of ex and constant on both sides, we get
3A + 3B = -12 ……(1)
and -4A = 20
∴ A = -5
from (1), 3(-5) + 3B = -12
∴ 3B = 3
∴ B = 1
∴ 20 – 12e^x = -5(3e^x – 4) + (3e^x)

Question 3.
\(\int \frac{3 e^{x}+4}{2 e^{x}-8} d x\)
Solution:
Let I = \(\int \frac{3 e^{x}+4}{2 e^{x}-8} d x\)
Put, Numerator = A (Denominator) + B[\(\frac{d}{d x}\)(Denominator)]
∴ 3e^x + 4 = A(2e^x – 8) + B[\(\frac{d}{d x}\)(2e^x – 8)]
∴ 3e^x + 4 = A(2e^x – 8) + B(2e^x – 0)
∴ 3e^x + 4 = (2A + 2B)e^x – 8A
Equating the coefficient of ex and constant on both sides, we get
2A + 2B = 3 ……..(1)
and -8A = 4
∴ A = \(-\frac{1}{2}\)
∴ from (1), 2(\(-\frac{1}{2}\)) + 2B = 3
∴ 2B = 4
∴ B = 2

Question 4.
\(\int \frac{2 e^{x}+5}{2 e^{x}+1} d x\)
Solution: