Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 2 Integral Calculus I Ex 2.7 Text Book Back Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.7

Question 1.
Integrate the following with respect to x.
$$\frac { 1 }{9-16x^2}$$
Solution:

Question 2.
$$\frac { 1} {9-8x-x^2}$$
Solution:

By Completing the squares
9 – 8x – x²
= -(x² + 8x – 9)
= – [x² + 8x + (4)² – (4)² – 9]
= – [(x + 4)² – 25]
= [25 – (x + 4)²]
= (5)² – (x + 4)²

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

Question 4.
$$\frac { 1 }{x^2-x-2}$$
Solution:

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

Question 6.
$$\frac { 1 }{2x^2+6x-8}$$
Solution:

Question 7.
$$\frac { e^x }{e^2x-9}$$
Solution:

Question 8.
$$\frac { 1 }{\sqrt {9x^2-7}}$$
Solution:

Question 9.
$$\frac { 1 }{\sqrt {x^2+16x+13}}$$
Solution:
∫$$\frac { 1 }{\sqrt {x^2+16x+13}}$$ dx
= ∫$$\frac { 1 }{\sqrt {(x+3)^2+(2)^2}}$$ dx
= log |(x + 3) + $$\sqrt {(x+3)^2+(2)^2}$$| + c
= log |(x + 3) + $$\sqrt {x^2+16x+13}$$| + c
By Completing the squares
x² + 6x + 3 = x² + 6x + (3)² – (3)² + 13
= (x + 3)² – 9 + 13
= (x + 3)² + 4
= (x + 3)² + (2)²

Question 10.
$$\frac { 1 }{ \sqrt x^2-3x+2 }$$
Solution:

Question 11.
$$\frac { x^3 }{ \sqrt x^8-1 }$$
Solution:

Question 12.
$$\sqrt { 1 + x + x^2}$$
Solution:

Question 13.
$$\sqrt { x^2 -2}$$
Solution:

Question 14.
$$\sqrt { 4x^2 -5}$$
Solution:

Question 15.
$$\sqrt { 2x^2 +4x+1}$$
Solution:

Question 16.
$$\frac { 1 }{ x + \sqrt x^2-1 }$$
Solution: