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.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter