Samacheer Kalvi 12th Business Maths Guide Chapter 2 Integral Calculus I Ex 2.7

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: