Let I(x) = 3dx(4x+6) 4x^2+8x+3 and I(0) = 34 + 20. If I ( 12) = a 2b + c, where a, b, c N, (a, b) = 1, then a+b+c is equal to

Given: I(x) = 3dx(4x+6) 4x^2+8x+3 and I(0) = 34 + 20. Find: a+b+c when I ( 12) = a 2b + c Rewrite the expression inside the square root: 4x^2+8x+3 = (2x+2)^2 - 1 Now take 2x+2 = Then 2dx = d so dx = 12 d Also, 4x+6 = 2(2x+2)+2 = 2 +2 = 2( +1) and 4x^2+8x+3 = ^2 -1 = Substituting in the integral, I = 3 ( 12 d)2( +1) Therefore, I = 34 +1 d Using +1 = 11+, I = 34 11+ d Using the half-angle identity 1+ = 2 ^2 ( 2) we get I = 34 12 ^2 ( 2) d so I = 38 ^2 ( 2) d Hence, I = 38 [2 ( 2) ] + C = 34 ( 2) + C Now convert back to x: ( 2) = 1- 1+ = -1 +1 = 2x+12x+3 Thus, I(x) = 34 2x+12x+3 + C Use the condition I(0) = 34 + 20. At x=0, 2x+12x+3 = 13 = 1 3 So, 34 1 3 + C = 34 + 20 Since 34 3 = 34 we obtain C = 20 Now evaluate at x= 12: 2x+12x+3 = 24 = 1 2 Hence, I ( 12) = 34 2 + 20 = 3 28 + 20 So, a=3, b=8, c=20 Therefore, a+b+c = 3+8+20 = 31 The correct option is D. Given: 4x^2+8x+3 = (2x+2)^2-1 Find: a substitution that turns the square root into a standard form. Because the expression is of the form u^2-1 with u=2x+2, the substitution u = is natural. This works because u^2-1 = which cancels neatly with the derivative of. After substitution, the integral reduces to I = 34 11+ d and the half-angle identity immediately gives I = 34 ( 2) + C Then use ( 2) = 2x+12x+3 to return to x. Applying the condition at x=0 gives C=20, and then at x= 12, I ( 12) = 3 28 + 20 Thus a=3, b=8, c=20 and a+b+c=31 So the correct option is D.

Let I(x) = 3dx(4x+6) 4x^2+8x+3 and I(0) = 34 + 20. If I ( 12) = a… — Solution | JEE Mathematics

MCQMediumJEE 2026Indefinite Integrals

Let $I(x) = \int \frac{3dx}{(4x+6)\sqrt{4x^2+8x+3}}$ and $I(0) = \frac{\sqrt{3}}{4} + 20$ . If $I\left(\frac{1}{2}\right) = \frac{a\sqrt{2}}{b} + c$ , where $a, b, c \in \mathbb{N}, \gcd(a, b) = 1$ , then $a+b+c$ is equal to

A
$29$
B
$28$
C
$30$
D
$31$

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given:

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

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

Find: $a+b+c$ when

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

Rewrite the expression inside the square root:

4x^2+8x+3 = (2x+2)^2 - 1

Now take

2x+2 = \sec\theta

Then

2dx = \sec\theta\tan\theta \, d\theta

dx = \frac{1}{2}\sec\theta\tan\theta \, d\theta

Also,

4x+6 = 2(2x+2)+2 = 2\sec\theta+2 = 2(\sec\theta+1)

and

\sqrt{4x^2+8x+3} = \sqrt{\sec^2\theta-1} = \tan\theta

Substituting in the integral,

I = \int \frac{3\left(\frac{1}{2}\sec\theta\tan\theta \, d\theta\right)}{2(\sec\theta+1)\tan\theta}

Therefore,

I = \frac{3}{4} \int \frac{\sec\theta}{\sec\theta+1} \, d\theta

Using $\frac{\sec\theta}{\sec\theta+1} = \frac{1}{1+\cos\theta}$ ,

I = \frac{3}{4} \int \frac{1}{1+\cos\theta} \, d\theta

Using the half-angle identity

1+\cos\theta = 2\cos^2\left(\frac{\theta}{2}\right)

we get

I = \frac{3}{4} \int \frac{1}{2\cos^2\left(\frac{\theta}{2}\right)} \, d\theta

I = \frac{3}{8} \int \sec^2\left(\frac{\theta}{2}\right) \, d\theta

Hence,

I = \frac{3}{8}\left[2\tan\left(\frac{\theta}{2}\right)\right] + C = \frac{3}{4}\tan\left(\frac{\theta}{2}\right) + C

Now convert back to $x$ :

\tan\left(\frac{\theta}{2}\right) = \sqrt{\frac{1-\cos\theta}{1+\cos\theta}} = \sqrt{\frac{\sec\theta-1}{\sec\theta+1}} = \sqrt{\frac{2x+1}{2x+3}}

Thus,

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

Use the condition $I(0) = \frac{\sqrt{3}}{4} + 20$ . At $x=0$ ,

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

So,

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

Since

\frac{3}{4\sqrt{3}} = \frac{\sqrt{3}}{4}

we obtain

C = 20

Now evaluate at $x=\frac{1}{2}$ :

\sqrt{\frac{2x+1}{2x+3}} = \sqrt{\frac{2}{4}} = \frac{1}{\sqrt{2}}

Hence,

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

So,

a=3, \quad b=8, \quad c=20

Therefore,

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

The correct option is D.

Trigonometric Substitution Insight

Given:

4x^2+8x+3 = (2x+2)^2-1

Find: a substitution that turns the square root into a standard form.

Because the expression is of the form $u^2-1$ with $u=2x+2$ , the substitution

u = \sec\theta

is natural. This works because

\sqrt{u^2-1} = \tan\theta

which cancels neatly with the derivative of $\sec\theta$ .

After substitution, the integral reduces to

I = \frac{3}{4}\int \frac{1}{1+\cos\theta} \, d\theta

and the half-angle identity immediately gives

I = \frac{3}{4}\tan\left(\frac{\theta}{2}\right) + C

Then use

\tan\left(\frac{\theta}{2}\right) = \sqrt{\frac{2x+1}{2x+3}}

to return to $x$ .

Applying the condition at $x=0$ gives $C=20$ , and then at $x=\frac{1}{2}$ ,

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

Thus $a=3$ , $b=8$ , $c=20$ and

a+b+c=31

So the correct option is D.

Common mistakes

Using an incorrect rewrite of the quadratic inside the root. The correct identity is $4x^2+8x+3 = (2x+2)^2-1$ , not $(2x+1)^2$ or any incomplete square. Always complete the square first before choosing a substitution.
Forgetting to transform $4x+6$ correctly after putting $2x+2 = \sec\theta$ . Since $4x+6 = 2(2x+2)+2$ , it becomes $2(\sec\theta+1)$ . Missing this factor changes the whole integrand.
Using the half-angle identity incorrectly. From $1+\cos\theta = 2\cos^2\left(\frac{\theta}{2}\right)$ , we get $\frac{1}{1+\cos\theta} = \frac{1}{2}\sec^2\left(\frac{\theta}{2}\right)$ . Do not replace it by $\sec^2\theta$ or miss the factor $\frac{1}{2}$ .
Making an error while converting $\tan\left(\frac{\theta}{2}\right)$ back to $x$ . Here $\sec\theta = 2x+2$ , so $\tan\left(\frac{\theta}{2}\right) = \sqrt{\frac{\sec\theta-1}{\sec\theta+1}} = \sqrt{\frac{2x+1}{2x+3}}$ . Substituting $x=\frac{1}{2}$ incorrectly leads to the wrong final value.

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