Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Indefinite Integration

Question:

Evaluate $\int \frac{3x-1}{\sqrt{x^2+9}} dx$.

Options:

$3\sqrt{x^2 + 9} - \ln|x + \sqrt{x^2 + 9}| + C$

$3\sqrt{x^2 + 9} + \ln|x + \sqrt{x^2 + 9}| + C$

$\frac{3}{2}\sqrt{x^2 + 9} - \ln|x + \sqrt{x^2 + 9}| + C$

$3\sqrt{x^2 + 9} - \arctan\left(\frac{x}{3}\right) + C$

Correct Answer:

$3\sqrt{x^2 + 9} - \ln|x + \sqrt{x^2 + 9}| + C$

Explanation:

The correct answer is Option (1) → $3\sqrt{x^2 + 9} - \ln|x + \sqrt{x^2 + 9}| + C$

Let $I = \int \frac{3x - 1}{\sqrt{x^2 + 9}} dx$

$\Rightarrow I = \int \frac{3x}{\sqrt{x^2 + 9}} dx - \int \frac{1}{\sqrt{x^2 + 9}} dx$

Let $I = I_1 - I_2$

Now, $I_1 = \int \frac{3x}{\sqrt{x^2 + 9}} dx$

Put $x^2 + 9 = t^2 \Rightarrow 2x \, dx = 2t \, dt \Rightarrow x \, dx = t \, dt$

$∴I_1 = 3 \int \frac{t}{t} dt$

$= 3 \int dt = 3t + C_1 = 3\sqrt{x^2 + 9} + C_1$

and

$I_2 = \int \frac{1}{\sqrt{x^2 + 9}} dx = \int \frac{1}{\sqrt{x^2 + (3)^2}} dx$

$= \log |x + \sqrt{x^2 + 9}| + C_2$

$∴I = 3\sqrt{x^2 + 9} + C_1 - \log |x + \sqrt{x^2 + 9}| - C_2$

$= 3\sqrt{x^2 + 9} - \log |x + \sqrt{x^2 + 9}| + C \quad [∵C = C_1 - C_2]$