Correct Answer is : 2x+34x2+3x−98log⁡(x+x2+3x)+c2 x+3/4 {x²+3 x}-9/8 ≤ft(x+{x²+3 x})+c42x+3x2+3x−89log(x+x2+3x)+c, where c is the constant of integration.

{x²+3 x}+ {x²+3 x}+c, where c is the constant of integration.

2 x+3/4 {x²+3 x}-9/8 ≤ft(x+{x²+3 x})+c, where c is the constant of integration.

x {x²+3 x}+ ≤ft(x+{x²+3 x})+c, where c is the constant of integration.

x+3 {x²+3 x}+3/2 ≤ft(x+{x²+3 x})+c, where c is the constant of integration.

Correct Answer is : 2x+34x2+3x−98log⁡(x+x2+3x)+c2 x+3/4 {x²+3 x}-9/8 ≤ft(x+{x²+3 x})+c42x+3x2+3x−89log(x+x2+3x)+c, where c is the constant of integration.

Test Index

MHTCET 23 Apr 2025 Shift 2 Paper

Question : 137 of 150

Marks: +1, -0

\int \sqrt{x^{2}+3 x} d x=

$\sqrt{x^{2}+3 x}+\log \sqrt{x^{2}+3 x}+c$ , where c is the constant of integration.
$\frac{2 x+3}{4} \sqrt{x^{2}+3 x}-\frac{9}{8} \log \left(x+\sqrt{x^{2}+3 x}\right)+c$ , where c is the constant of integration.
$x \sqrt{x^{2}+3 x}+\log \left(x+\sqrt{x^{2}+3 x}\right)+c$ , where c is the constant of integration.
$x+3 \sqrt{x^{2}+3 x}+\frac{3}{2} \log \left(x+\sqrt{x^{2}+3 x}\right)+c$ , where c is the constant of integration.

Validate

Solution:

Concept:

Use the method of completing the square and the standard integral formula for

\int \sqrt{t^{2}-a^{2}} \, dt

Explanation:

Let

I = \int \sqrt{x^{2}+3x} \, dx

Complete the square inside the root:

x^{2}+3x = x^{2}+3x+\frac{9}{4}-\frac{9}{4} = \left(x+\frac{3}{2}\right)^{2}-\left(\frac{3}{2}\right)^{2}

I = \int \sqrt{\left(x+\frac{3}{2}\right)^{2}-\left(\frac{3}{2}\right)^{2}} \, dx

Using the formula

\int \sqrt{t^{2}-a^{2}} \, dt = \frac{t}{2}\sqrt{t^{2}-a^{2}} - \frac{a^{2}}{2}\log\left|t+\sqrt{t^{2}-a^{2}}\right| + c

take

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

and

a = \frac{3}{2}

Then

I = \frac{x+\frac{3}{2}}{2}\sqrt{\left(x+\frac{3}{2}\right)^{2}-\left(\frac{3}{2}\right)^{2}} - \frac{\left(\frac{3}{2}\right)^{2}}{2}\log\left(x+\frac{3}{2}+\sqrt{\left(x+\frac{3}{2}\right)^{2}-\left(\frac{3}{2}\right)^{2}}\right)+c

Simplify using

\left(x+\frac{3}{2}\right)^{2}-\left(\frac{3}{2}\right)^{2} = x^{2}+3x

I = \frac{2x+3}{4}\sqrt{x^{2}+3x} - \frac{9}{8}\log\left(x+\frac{3}{2}+\sqrt{x^{2}+3x}\right)+c

This matches option B.

Answer:

\frac{2x+3}{4}\sqrt{x^{2}+3x} - \frac{9}{8}\log\left(x+\sqrt{x^{2}+3x}\right)+c

(Option B), where

c

is the constant of integration.

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