Mathematics Questions and Answers – Integrals of Some Particular Functions

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This set of Mathematics Exam Questions for IIT JEE Exam focuses on “Integrals of Some Particular Functions”.

1. Find \(\int \frac{2 dx}{x^2-64}\).
a) –\(log⁡\left |\frac{x+8}{x-8}\right |+C\)
b) \(\frac{3}{2} log⁡\left |\frac{x+8}{x-8}\right |+C\)
c) \(log⁡\left |\frac{x+8}{x-8}\right |+C\)
d) \(\frac{1}{8} log⁡\left |\frac{x-8}{x+8}\right |+C\)
View Answer

Answer: d
Explanation: \(\int \frac{2 dx}{x^2-64}=2\int \frac{dx}{x^2-8^2}\)
By using the formula \(\int \frac{dx}{x^2-a^2}=\frac{1}{2a} log⁡|\frac{x-a}{x+a}|+C\)
∴\(2\int \frac{dx}{x^2-8^2}=2(\frac{1}{(2(8))} log⁡|\frac{x-8}{x+8}|)+2C_1\)
=\(\frac{1}{8} log⁡|\frac{x-8}{x+8}|+C\)
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2. Find \(\int \frac{8 dx}{x^2-16}\).
a) \(log⁡\left |\frac{4+x}{4-x}\right |+C\)
b) –\(log⁡\left |\frac{4+x}{4-x}\right |+C\)
c) \(8 log⁡\left |\frac{4+x}{4-x}\right |+C\)
d) \(\frac{1}{8} log⁡\left |\frac{4+x}{4-x}\right |+C\)
View Answer

Answer: a
Explanation: \(\int \frac{8dx}{16-x^2}=8\int \frac{dx}{4^2-x^2}\)
By using the formula \(\int \frac{dx}{a^2-x^2}=\frac{1}{2a} \left |\frac{a+x}{a-x}\right |+C\)
∴\(8\int \frac{dx}{4^2-x^2}=8(\frac{1}{2(4)} log⁡\left |\frac{4+x}{4-x}\right |)+8C_1\)
\(8\int \frac{dx}{4^2-x^2}=log⁡\left |\frac{4+x}{4-x}\right |+C\)

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) \(tan^{-1}⁡\frac{x}{4}+C\)
View Answer

Answer: b
Explanation: \(\int \frac{3dx}{9+x^2}=3\int \frac{dx}{3^2+x^2}\)
Using the formula \(\int \frac{dx}{a^2+x^2}=\frac{1}{a} tan^{-1}\frac{⁡x}{a}+C\)
∴\(3\int \frac{dx}{x^2+3^2}=3\left (\frac{1}{3} tan^{-1}⁡\frac{x}{3}\right )+3C_1\)
\(3\int \frac{dx}{x^2+3^2}=tan^{-1}⁡\frac{x}{3}+C\).
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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 \(log⁡|x+\sqrt{x^2-25}|+C\)
View Answer

Answer: c
Explanation: \(\int \frac{10 \,dx}{\sqrt{x^2-25}}=10\int \frac{dx}{\sqrt{x^2-25}}\)
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) \(sin^{-1}⁡\frac{x}{5}+C\)
View Answer

Answer: a
Explanation: \(\int \frac{dx}{\sqrt{5-x^2}}=\int \frac{dx}{\sqrt{(√5)^2-x^2}}\)
By using the formula \(\int \frac{dx}{\sqrt{a^2-x^2}}=sin^{-1}⁡\frac{x}{a}+C\)
∴\(\int \frac{dx}{\sqrt{x^2-5}}=sin^{-1}⁡\frac{x}{\sqrt{5}}+C\)
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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) \(log⁡|x^2+\sqrt{x^2+36}|+C\)
View Answer

Answer: d
Explanation: \(\int \frac{dx}{\sqrt{x^2+36}}\)
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) \(x-\frac{1}{2} tan^{-1}⁡\frac{x-4}{2}+C\)
View Answer

Answer: c
Explanation: \(\int \frac{dx}{x^2-8x+20}=\int \frac{dx}{(x^2-2(4x)+4^2)+4}\)
=\(\int \frac{dx}{(x-4)^2+2^2}\)
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
\(\int \frac{dx}{(x-4)^2+2^2}=\frac{1}{2} tan^{-1}⁡\frac{x-4}{2}+C\)
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8. Find \(\int \frac{(x+3)}{2x^2+6x+7} dx\).
a) \(\frac{1}{4} log⁡(2x^2+6x+7) + \frac{3}{4} \left (\frac{1}{\sqrt{2}} tan^{-1}⁡\frac{2x+3}{2\sqrt{2}}\right )+C\)
b) \(\frac{1}{4} log⁡(2x^2+6x+7) – \frac{3}{4} (\frac{1}{\sqrt{2}} tan^{-1}⁡\frac{2x+3}{2\sqrt{2}} )+C\)
c) \(log⁡(2x^2+6x+7) + \left (tan^{-1}⁡\frac{2x+3}{2√2}\right )+C\)
d) –\(log⁡(2x^2+6x+7) – \frac{3}{4} \left (\frac{1}{√2} tan^{-1}⁡\frac{2x+3}{2√2}\right )+C\)
View Answer

Answer: a
Explanation: 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
\(\frac{1}{4} \int \frac{4x+6}{2x^2+6x+7} dx=\frac{1}{4} \int \frac{dt}{t}=\frac{1}{4} log⁡t\)
Replacing t with (2x2+6x+7)
\(\frac{1}{4} \int \frac{4x+6}{2x^2+6x+7} dx=\frac{1}{4} log⁡(2x^2+6x+7)\)
\(\frac{3}{2} \int \frac{1}{2x^2+6x+7} dx=\frac{3}{2} \int \frac{1}{2(x^2+3x+\frac{7}{2})} dx=\frac{3}{4} \int \frac{1}{(x+\frac{3}{2})^2+2} dx\)
=\(\frac{3}{4} \left (\frac{1}{\sqrt{2}} tan^{-1}⁡\frac{2x+3}{2\sqrt{2}} \right )\)
∴\(\int \frac{x+3}{2x^2+6x+7} dx=\frac{1}{4} log⁡(2x^2+6x+7) + \frac{3}{4} \left (\frac{1}{\sqrt{2}} tan^{-1}⁡\frac{2x+3}{2√2} \right )+C\)

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) \(\frac{7}{6} log⁡|\frac{x-3}{x+3}|+C\)
View Answer

Answer: d
Explanation: \(\int \frac{7dx}{x^2-9}=2\int \frac{7dx}{x^2-3^2}\)
By using the formula \(\int \frac{dx}{x^2-a^2}=\frac{1}{2a} log⁡|\frac{x-a}{x+a}|+C\)
∴\(7\int \frac{dx}{x^2-3^2}=7(\frac{1}{2(3)} log⁡|\frac{x-3}{x+3}|)+7C_1\)
=\(\frac{7}{6} log⁡|\frac{x-3}{x+3}|+C\)
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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) \(\frac{3}{4} tan^{-1}\frac{⁡3x}{2}+C\)
View Answer

Answer: b
Explanation: \(\int \frac{dx}{x^2+4}=\int \frac{dx}{x^2+2^2}\)
Using the formula \(\int \frac{dx}{a^2+x^2}=\frac{1}{a} tan^{-1}⁡\frac{x}{a}+C\)
∴\(\int \frac{dx}{x^2+2^2}=(\frac{1}{2} tan^{-1}⁡\frac{x}{2})+C\)

Sanfoundry Global Education & Learning Series – Mathematics – Class 12.

To practice Mathematics Exam Questions for IIT JEE Exam, here is complete set of 1000+ Multiple Choice Questions and Answers.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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