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Indefinite Integrals

Section: Mathematics

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Question : 5 of 90

Marks: +1, -0

Let

I (x)=\int\;\frac{3\,dx}{(4x+6)\left(\sqrt{4x^2+8x+3}\right)}

I (0)=\;\frac{\sqrt{3}}{4}+20

I\left(\frac{1}{2}\right)=\;\frac{a\sqrt{2}}{b}+c

a, b, c \in N , \gcd(a,b)=1

a+b+c

[23 jan 2026 Shift 2]

30
29
28
31

Solution:

Concept:

The integral is solved by rewriting the expression and using a substitution to simplify the square root.

Explanation:

First, factor the integrand:

\int \frac{3\,dx}{(4x+6)\sqrt{4x^2+8x+3}} = \int \frac{3\,dx}{2(2x+3)\sqrt{(2x+2)^2-1}}

.

Let

2x+3 = \frac{1}{t}

. Then

2\,dx = -\frac{1}{t^2}\,dt

, so

dx = -\frac{1}{2t^2}\,dt

. Also

2x+2 = \frac{1}{t} - 1 = \frac{1-t}{t}

.

Substitute:

I(x) = \int \frac{3\left(-\frac{1}{2t^2}dt\right)}{2\left(\frac{1}{t}\right)\sqrt{\left(\frac{1-t}{t}\right)^2-1}} = \int \frac{-3\,dt}{4t \sqrt{\frac{1-2t+t^2-t^2}{t^2}}} = \int \frac{-3\,dt}{4\sqrt{1-2t}}

.

Now integrate:

I(x) = -\frac{3}{4} \int (1-2t)^{-1/2}\,dt = -\frac{3}{4} \left[ \frac{(1-2t)^{1/2}}{(1/2)(-2)} \right] + C = \frac{3}{4}\sqrt{1-2t} + C

.

Substitute back

t = \frac{1}{2x+3}

:

I(x) = \frac{3}{4}\sqrt{1-\frac{2}{2x+3}} + C = \frac{3}{4}\sqrt{\frac{2x+1}{2x+3}} + C

.

Use

I(0) = \frac{\sqrt{3}}{4} + 20

:

\frac{3}{4}\sqrt{\frac{1}{3}} + C = \frac{\sqrt{3}}{4} + 20 \Rightarrow \frac{\sqrt{3}}{4} + C = \frac{\sqrt{3}}{4} + 20 \Rightarrow C = 20

.

Thus

I(x) = \frac{3}{4}\sqrt{\frac{2x+1}{2x+3}} + 20

.

Find

I\left(\frac{1}{2}\right)

:

I\left(\frac{1}{2}\right) = \frac{3}{4}\sqrt{\frac{2(1/2)+1}{2(1/2)+3}} + 20 = \frac{3}{4}\sqrt{\frac{2}{4}} + 20 = \frac{3\sqrt{2}}{8} + 20

.

Compare with

\frac{a\sqrt{2}}{b} + c

:

a=3

,

b=8

,

c=20

.

Then

a+b+c = 3+8+20 = 31

.

Answer:

31

, which corresponds to option D.

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