Mathematics Exam Questions for IIT JEE Exam on “Integrals of Some Particular Functions”.
1. Find (int frac{2 dx}{x^2-64}).
a) –(logleft |frac{x+8}{x-8}right |+C)
b) (frac{3}{2} logleft |frac{x+8}{x-8}right |+C)
c) (logleft |frac{x+8}{x-8}right |+C)
d) (frac{1}{8} logleft |frac{x-8}{x+8}right |+C)
Answer: d
Clarification: (int frac{2 dx}{x^2-64}=2int frac{dx}{x^2-8^2})
By using the formula (int frac{dx}{x^2-a^2}=frac{1}{2a} log|frac{x-a}{x+a}|+C)
∴(2int frac{dx}{x^2-8^2}=2(frac{1}{(2(8))} log|frac{x-8}{x+8}|)+2C_1)
=(frac{1}{8} log|frac{x-8}{x+8}|+C)
2. Find (int frac{8 dx}{x^2-16}).
a) (logleft |frac{4+x}{4-x}right |+C)
b) –(logleft |frac{4+x}{4-x}right |+C)
c) (8 logleft |frac{4+x}{4-x}right |+C)
d) (frac{1}{8} logleft |frac{4+x}{4-x}right |+C)
Answer: a
Clarification: (int frac{8dx}{16-x^2}=8int frac{dx}{4^2-x^2})
By using the formula (int frac{dx}{a^2-x^2}=frac{1}{2a} left |frac{a+x}{a-x}right |+C)
∴(8int frac{dx}{4^2-x^2}=8(frac{1}{2(4)} logleft |frac{4+x}{4-x}right |)+8C_1)
(8int frac{dx}{4^2-x^2}=logleft |frac{4+x}{4-x}right |+C)
3. Find (int frac{3dx}{9+x^2}).
a) (tan^{-1}frac{x}{2}+C)
b) (tan^{-1}frac{x}{3}+C)
c) (tan^{-1}frac{x}{5}+C)
d) (tan^{-1}frac{x}{4}+C)
Answer: b
Clarification: (int frac{3dx}{9+x^2}=3int frac{dx}{3^2+x^2})
Using the formula (int frac{dx}{a^2+x^2}=frac{1}{a} tan^{-1}frac{x}{a}+C)
∴(3int frac{dx}{x^2+3^2}=3left (frac{1}{3} tan^{-1}frac{x}{3}right )+3C_1)
(3int frac{dx}{x^2+3^2}=tan^{-1}frac{x}{3}+C).
4. Find (int frac{10 ,dx}{sqrt{x^2-25}}).
a) –(log|x+sqrt{x^2-25}|+C)
b) (log|x+sqrt{x^2-25}|+C)
c) 10 ( log|x+sqrt{x^2-25}|+C)
d) -10 (log|x+sqrt{x^2-25}|+C)
Answer: c
Clarification: (int frac{10 ,dx}{sqrt{x^2-25}}=10int frac{dx}{sqrt{x^2-25}})
By using the formula (int frac{dx}{sqrt{x^2-a^2}}=log|x+sqrt{x^2-a^2}|+C), we get
∴(10 int frac{dx}{sqrt{x^2-25}}=10 log|x+sqrt{x^2-25}|+10C_1)
=10 (log|x+sqrt{x^2-25}|+C)
5. Find (int frac{dx}{sqrt{5-x^2}}).
a) (sin^{-1}frac{x}{sqrt{5}}+C)
b) (2 sin^{-1}frac{x}{sqrt{5}}+C)
c) –(sin^{-1}frac{x}{sqrt{5}}+C)
d) (sin^{-1}frac{x}{5}+C)
Answer: a
Clarification: (int frac{dx}{sqrt{5-x^2}}=int frac{dx}{sqrt{(√5)^2-x^2}})
By using the formula (int frac{dx}{sqrt{a^2-x^2}}=sin^{-1}frac{x}{a}+C)
∴(int frac{dx}{sqrt{x^2-5}}=sin^{-1}frac{x}{sqrt{5}}+C)
6. Integrate (frac{dx}{sqrt{x^2+36}}).
a) –(log|x^2+sqrt{x^2+36}|+C)
b) (log|2x+sqrt{x^2+36}|+C)
c) –(log|x^2+sqrt{x^2+6}|+C)
d) (log|x^2+sqrt{x^2+36}|+C)
Answer: d
Clarification: (int frac{dx}{sqrt{x^2+36}})
By using the formula (int frac{dx}{sqrt{x^2+a^2}}=log|x^2+sqrt{x^2+a^2}|+C)
∴(int frac{dx}{sqrt{x^2+36}}=log|x^2+sqrt{x^2+36}|+C)
7. Find (int frac{dx}{x^2-8x+20}).
a) (frac{1}{2} tan^{-1}frac{x^2-8x}{2}+C)
b) (frac{5}{2} tan^{-1}frac{x-4}{2}+C)
c) (frac{1}{2} tan^{-1}frac{x-4}{2}+C)
d) (x-frac{1}{2} tan^{-1}frac{x-4}{2}+C)
Answer: c
Clarification: (int frac{dx}{x^2-8x+20}=int frac{dx}{(x^2-2(4x)+4^2)+4})
=(int frac{dx}{(x-4)^2+2^2})
Let x-4=t
Differentiating w.r.t x, we get
dx=dt
By using the formula (int frac{dx}{x^2+a^2}=frac{1}{a} tan^{-1}frac{x}{a}+C)
(int frac{dx}{(x-4)^2+2^2}=int frac{dt}{t^2+2^2}=frac{1}{2} tan^{-1}frac{t}{2}+C)
Replacing t with x-4, we get
(int frac{dx}{(x-4)^2+2^2}=frac{1}{2} tan^{-1}frac{x-4}{2}+C)
8. Find (int frac{(x+3)}{2x^2+6x+7} dx).
a) (frac{1}{4} log(2x^2+6x+7) + frac{3}{4} left (frac{1}{sqrt{2}} tan^{-1}frac{2x+3}{2sqrt{2}}right )+C)
b) (frac{1}{4} log(2x^2+6x+7) – frac{3}{4} (frac{1}{sqrt{2}} tan^{-1}frac{2x+3}{2sqrt{2}} )+C)
c) (log(2x^2+6x+7) + left (tan^{-1}frac{2x+3}{2√2}right )+C)
d) –(log(2x^2+6x+7) – frac{3}{4} left (frac{1}{√2} tan^{-1}frac{2x+3}{2√2}right )+C)
Answer: a
Clarification: We can express
x+3=A (frac{d}{dx}) (2x2+6x+7)+B
x+3=A(4x+6)+B
x+3=4Ax+(6A+B)
Comparing the coefficients, we get
4A=1 ⇒A=1/4
6A+B=3 ⇒B=3/2
(int frac{x+3}{2x^2+6x+7} dx=frac{1}{4} int frac{4x+6}{2x^2+6x+7} dx+frac{3}{2} int frac{1}{2x^2+6x+7} dx)
Let 2x2+6x+7=t
(4x+6)dx=dt
(frac{1}{4} int frac{4x+6}{2x^2+6x+7} dx=frac{1}{4} int frac{dt}{t}=frac{1}{4} logt)
Replacing t with (2x2+6x+7)
(frac{1}{4} int frac{4x+6}{2x^2+6x+7} dx=frac{1}{4} log(2x^2+6x+7))
(frac{3}{2} int frac{1}{2x^2+6x+7} dx=frac{3}{2} int frac{1}{2(x^2+3x+frac{7}{2})} dx=frac{3}{4} int frac{1}{(x+frac{3}{2})^2+2} dx)
=(frac{3}{4} left (frac{1}{sqrt{2}} tan^{-1}frac{2x+3}{2sqrt{2}} right ))
∴(int frac{x+3}{2x^2+6x+7} dx=frac{1}{4} log(2x^2+6x+7) + frac{3}{4} left (frac{1}{sqrt{2}} tan^{-1}frac{2x+3}{2√2} right )+C)
9. Find (int frac{7dx}{x^2-9}).
a) (frac{7}{6} log|frac{x-9}{x+9}|+C)
b) (frac{7}{9} log|frac{x-3}{x+3}|+C)
c) –(frac{7}{6} log|frac{x+3}{x-3}|+C)
d) (frac{7}{6} log|frac{x-3}{x+3}|+C)
Answer: d
Clarification: (int frac{7dx}{x^2-9}=2int frac{7dx}{x^2-3^2})
By using the formula (int frac{dx}{x^2-a^2}=frac{1}{2a} log|frac{x-a}{x+a}|+C)
∴(7int frac{dx}{x^2-3^2}=7(frac{1}{2(3)} log|frac{x-3}{x+3}|)+7C_1)
=(frac{7}{6} log|frac{x-3}{x+3}|+C)
10. Find (int frac{dx}{x^2+4}).
a) –(tan^{-1}frac{x}{4}+C)
b) (frac{1}{2} tan^{-1}frac{x}{2}+C)
c) (frac{3}{4} tan^{-1}x+C)
d) (frac{3}{4} tan^{-1}frac{3x}{2}+C)
Answer: b
Clarification: (int frac{dx}{x^2+4}=int frac{dx}{x^2+2^2})
Using the formula (int frac{dx}{a^2+x^2}=frac{1}{a} tan^{-1}frac{x}{a}+C)
∴(int frac{dx}{x^2+2^2}=(frac{1}{2} tan^{-1}frac{x}{2})+C)
Mathematics Exam Questions for IIT JEE Exam,