Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 2 Integral Calculus I Ex 2.6 Text Book Back Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.6

Question 1.
Integrate the following with respect to x.
$$\frac { 2x+5 }{x^2+5x-7}$$
Solution:
∫$$\frac { 2x+5 }{x^2+5x-7}$$ dx
∫$$\frac { 1 }{z}$$ dz
= log |z| + c
= log |x² + 5x – 7| + c
Take z = x² + 5x – 7
$$\frac { dz }{dx}$$ = 2x + 5
dz = (2x + 5)dx

Question 2.
$$\frac { e^{3logx} }{x^4+1}$$
Solution:

Question 3.
$$\frac { e^{2x} }{e^{2x}-2}$$
Solution:

Question 4.
$$\frac { (logx)^3 }{x}$$
Solution:

Question 5.
$$\frac { 6x+7 }{\sqrt{3x^2+7x-1}}$$
Solution:

Question 6.
(4x + 2) $$\sqrt {x^2+x+1}$$
Solution:

Question 7.
x8 (1 + x9)5
Solution:

Question 8.
$$\frac { x^{e-1}+e^{x-1} }{x^e+e^x}$$
Solution:

Question 9.
$$\frac { 1 }{x log x}$$
Solution:
∫$$\frac { 1 }{x log x}$$ dx
∫$$\frac { 1 }{z}$$ dz
= log |z| + c
= log |log x| + c
Take z = log x
$$\frac { 1 }{z}$$ = $$\frac { 1 }{x}$$
dz = $$\frac { 1 }{x}$$ dx

Question 10.
$$\frac { x }{2x^4-3x^2-2}$$
Solution:

Question 11.
ex (1 + x) log (xex)
Solution:
ex (1 + x) log(x ex) = (ex + x ex) log (x ex)
Let z = x ex, Then dz = d(x ex)
dz = (x ex + ex) dx (Using product rule)
So ∫ ex (1 + x) log (x ex) dx
= ∫ log (x ex) (ex + x ex) dx
= ∫ log z dz
= z (log z – 1) + c
= x ex [log (x ex) – 1] + c

Question 12.
$$\frac { 1 }{x(x^2+1)}$$
Solution:

Put x = 0
1 = A(1)
A = 1
Put x = 1
1 = A(2) + 1(B + C)
1 = (1)2 + B + C
B + C = -1 …….. (1)
Put x = -1
1 = A[(-1)² + 1] + (-1)[B(-1) + C]
1 = A(2) – (-B + C)
1 = 2A + B – C
1 = 2(1) + B – C
B – C = -1 ………. (2)
2B = -2
B = -1
C = 0

Question 13.
ex [ $$\frac { 1 }{x^2}$$ – $$\frac { 2 }{x^3}$$ ]
Solution:
∫ex [ $$\frac { 1 }{x^2}$$ – $$\frac { 2 }{x^3}$$ ] dx
= ∫ex [f(x) + f'(x)] dx
= ex f(x) + c
= ex [ $$\frac { 1 }{x^2}$$ ] + c
Take
f(x) = $$\frac { 1 }{x^2}$$
f'(x) = $$\frac { -2 }{x^3}$$

Question 14.
ex [ $$\frac { x-1 }{(x+1)^3}$$ ]
Solution:

Question 15.
e3x [ $$\frac { 3x-1 }{9x^2}$$ ]
Solution: