# 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) $$\frac{1}{8} log⁡\left |\frac{x-8}{x+8}\right |+C$$

Explanation: $$\int \frac{2 dx}{x^2-64}=2\int \frac{dx}{x^2-8^2}$$
By using the formula $$\int \frac{dx}{x^2-a^2}=\frac{1}{2a} log⁡|\frac{x-a}{x+a}|+C$$
∴$$2\int \frac{dx}{x^2-8^2}=2(\frac{1}{(2(8))} log⁡|\frac{x-8}{x+8}|)+2C_1$$
=$$\frac{1}{8} log⁡|\frac{x-8}{x+8}|+C$$

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) $$\frac{1}{8} log⁡\left |\frac{4+x}{4-x}\right |+C$$

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) $$tan^{-1}⁡\frac{x}{4}+C$$

Explanation: $$\int \frac{3dx}{9+x^2}=3\int \frac{dx}{3^2+x^2}$$
Using the formula $$\int \frac{dx}{a^2+x^2}=\frac{1}{a} tan^{-1}\frac{⁡x}{a}+C$$
∴$$3\int \frac{dx}{x^2+3^2}=3\left (\frac{1}{3} tan^{-1}⁡\frac{x}{3}\right )+3C_1$$
$$3\int \frac{dx}{x^2+3^2}=tan^{-1}⁡\frac{x}{3}+C$$.

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 $$log⁡|x+\sqrt{x^2-25}|+C$$

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) $$sin^{-1}⁡\frac{x}{5}+C$$

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

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) $$log⁡|x^2+\sqrt{x^2+36}|+C$$

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) $$x-\frac{1}{2} tan^{-1}⁡\frac{x-4}{2}+C$$

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
$$\int \frac{dx}{(x-4)^2+2^2}=\frac{1}{2} tan^{-1}⁡\frac{x-4}{2}+C$$

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) –$$log⁡(2x^2+6x+7) – \frac{3}{4} \left (\frac{1}{√2} tan^{-1}⁡\frac{2x+3}{2√2}\right )+C$$

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) $$\frac{7}{6} log⁡|\frac{x-3}{x+3}|+C$$

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

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) $$\frac{3}{4} tan^{-1}\frac{⁡3x}{2}+C$$

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