OP Malhotra Class 12 Maths Solutions Chapter 15 Indefinite Integral-3 Ex 15(e)

Well-structured ISC Class 12 Maths Solutions OP Malhotra Chapter 15 Indefinite Integral-3 Ex 15(e) facilitate a deeper understanding of mathematical principles.

S Chand Class 12 ICSE Maths Solutions Chapter 15 Indefinite Integral-3 Ex 15(e)

Question 1.
\(\int \frac{d x}{\sqrt{7-6 x-x^2}}\)
Solution:
Let I = \(\int \frac{d x}{\sqrt{7-6 x-x^2}}\)
= \(\int \frac{d x}{\sqrt{-\left(x^2+6 x-7\right)}}\)
= \(\int \frac{d x}{\sqrt{-\left(x^2+6 x+9-16\right)}}\)
= \(\int \frac{d x}{\sqrt{16-(x+3)^2}}\)
Put x + 3 = t ⇒ dx = dt
= \(\int \frac{d t}{\sqrt{4^2-t^2}}\) = sin^-1\(\left(\frac{t}{4}\right)\) + C
= sin^-1\(\left(\frac{x+3}{4}\right)\) + C

Question 2.
\(\int \frac{d x}{\sqrt{10-8 x-2 x^2}}\)
Solution:
Let I = \(\int \frac{d x}{\sqrt{10-8 x-2 x^2}}\)
= \(\int \frac{d x}{\sqrt{-2\left(x^2+4 x-5\right)}}\)
= \(\frac{1}{\sqrt{2}}\)\(\int \frac{d x}{\sqrt{-\left(x^2+4 x+4-9\right)}}\)
= \(\frac{1}{\sqrt{2}}\)\(\int \frac{d x}{\sqrt{9-(x+2)^2}}\)
Put x + 2 = t ⇒ dx = dt
= \(\frac{1}{\sqrt{2}}\)\(\int \frac{d t}{\sqrt{3^2-t^2}}\)
= \(\frac{1}{\sqrt{2}}\)sin^-1\(\left(\frac{x+2}{3}\right)\) + C

Question 3.
\(\int \frac{d x}{\sqrt{4-2 x-2 x^2}}\)
Solution:

Question 4.
\(\int \frac{d x}{\sqrt{16-2 x-2 x^2}}\)
Solution:

Question 5.
\(\int \frac{e^x}{\sqrt{5-4 e^x-e^{2 x}}} d x\)
Solution:
Let I = \(\int \frac{e^x}{\sqrt{5-4 e^x-e^{2 x}}} d x\)
Put e^x = t ⇒ e^x dx = dt
= \(\int \frac{d t}{\sqrt{5-4 t-t^2}}\) = \(\int \frac{d t}{\sqrt{-\left(t^2+4 t-5\right)}}\)
= \(\int \frac{d x}{\sqrt{-\left(t^2+4 t+4-9\right)}}\)
= \(\int \frac{d x}{\sqrt{-\left\{(t+2)^2-9\right\}}}\)
= \(\int \frac{d t}{\sqrt{3^2-(t+2)^2}}\)
Put t + 2 = u ⇒ dt = du
= \(\int \frac{d u}{\sqrt{3^2-u^2}}\) = sin^-1\(\left(\frac{u}{3}\right)\) + C
= sin^-1\(\left(\frac{t+2}{3}\right)\) + C

Question 6.
\(\int \frac{1}{\sqrt{(x-1)(x-2)}} d x\)
Solution:

Question 7.
\(\int \frac{1}{\sqrt{5 x^2-2 x}} d x\)
Solution:

Question 8.
\(\int \frac{2 x+1}{\sqrt{x^2+2 x-1}} d x\)
Solution:
Let I = \(\int \frac{2 x+1}{\sqrt{x^2+2 x-1}} d x\)
= \(\int \frac{2 x+2-1}{\sqrt{x^2+2 x-1}} d x\)
= \(\int \frac{(2 x+2) d x}{\sqrt{x^2+2 x-1}}\) – \(\int \frac{d x}{\sqrt{x^2+2 x-1}}\)
Put x² + 2x – 1 = t in first integral ⇒ (2x + 2)dx = dt

Question 9.
\(\int \frac{x d x}{\sqrt{8+x-x^2}}\)
Solution:
Let I = \(\int \frac{x d x}{\sqrt{8+x-x^2}}\) = \(\frac{1}{-2}\)\(\int \frac{(1-2 x-1) d x}{\sqrt{8+x-x^2}}\)
= –\(\frac{1}{2}\)\(\int \frac{(1-2 x) d x}{\sqrt{8+x-x^2}}\) + \(\frac{1}{2}\)\(\int \frac{d x}{\sqrt{-\left(x^2-x-8\right)}}\)
Put 8 + x – x² = t in first integral ⇒ (1 – 2x)dx = dt

Question 10.
\(\int \frac{6 x+7}{\sqrt{(x-5)(x-4)}} d x\)
Solution:

Question 11.
\(\int \frac{4 x+1}{\sqrt{2 x^2+x-3}} d x\)
Solution:
Let I = \(\int \frac{4 x+1}{\sqrt{2 x^2+x-3}} d x\)
Put 2x² + x – 3 = t ⇒ (4x + 1)dx = dt
= \(\int \frac{d t}{\sqrt{t}}\) = \(\int t^{-\frac{1}{2}} d t\) = \(\frac{t^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\) + C
= \(2 \sqrt{t}\) + C = \(2 \sqrt{2 x^2+x-3}\) + C

Question 12.
\(\int \frac{x}{\sqrt{x^2+x+1}} d x\)
Solution:
Let I = \(\int \frac{x}{\sqrt{x^2+x+1}} d x\)
= \(\frac { 1 }{ 2 }\)\(\int \frac{2 x+1-1 d x}{\sqrt{x^2+x+1}}\)
= \(\frac { 1 }{ 2 }\)\(\int \frac{2 x+1}{\sqrt{x^2+x+1}} d x\) – \(\frac { 1 }{ 2 }\)\(\int \frac{d x}{\sqrt{x^2+x+1}}\)
Put x² + x + 1 = t in first integral ⇒ (2x + 1)dx = dt

Question 13.
\(\int \frac{x+2}{\sqrt{x^2-1}} d x\)
Solution:
Let I = \(\int \frac{x+2}{\sqrt{x^2-1}} d x\)
= \(\int \frac{x}{\sqrt{x^2-1}} d x\) + 2\(\int \frac{d x}{\sqrt{x^2-1^2}}\)
Put x² – 1 = t ⇒ 2xdx = dt

Question 14.
\(\int \sqrt{\frac{5-x}{x-2}} d x\)
Solution:

Question 15.
\(\int \frac{5 x+3}{\sqrt{x^2+4 x+10}} d x\)
Solution:

Question 16.
\(\int \frac{x+2}{\sqrt{x^2+5 x+6}} d x\)
Solution: