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: