₀² {3 x+1}{x²+4} d x=

Correct Answer is : log⁡(22)+π8 (2 {2})+π/8log(22)+8π

Correct Answer is : log⁡(22)+π8 (2 {2})+π/8log(22)+8π

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MHTCET 22 Apr 2025 Shift 1 Paper

Question : 129 of 150

Marks: +1, -0

\int_{0}^{2} \frac{3 x+1}{x^{2}+4} d x=

Solution:

Concept:

Split the integrand and use standard formulas:

\int \frac{f'(x)}{f(x)}\,dx = \log|f(x)|

and

\int \frac{1}{x^2+a^2}\,dx = \frac{1}{a}\tan^{-1}\frac{x}{a}

Explanation:

Let

I = \int_{0}^{2} \frac{3x+1}{x^2+4}\,dx

Write it as

I = I_1 + I_2

, where

I_1 = \int_{0}^{2} \frac{3x}{x^2+4}\,dx

and

I_2 = \int_{0}^{2} \frac{1}{x^2+4}\,dx

For

I_1

, put

t = x^2+4

, so

dt = 2x\,dx

, hence

x\,dx = \frac{dt}{2}

When

x=0

t=4

; when

x=2

t=8

I_1 = \frac{3}{2} \int_{4}^{8} \frac{dt}{t} = \frac{3}{2}[\log t]_{4}^{8}

= \frac{3}{2}(\log 8 - \log 4) = \frac{3}{2}(3\log 2 - 2\log 2) = \frac{3}{2}\log 2 = \log(2\sqrt{2})

For

I_2

, use

\int \frac{dx}{x^2+4} = \frac{1}{2}\tan^{-1}\frac{x}{2}

I_2 = \frac{1}{2}[\tan^{-1}\frac{x}{2}]_{0}^{2} = \frac{1}{2}(\tan^{-1}1 - \tan^{-1}0) = \frac{1}{2} \cdot \frac{\pi}{4} = \frac{\pi}{8}

Therefore,

I = I_1 + I_2 = \log(2\sqrt{2}) + \frac{\pi}{8}

Answer:

\log(2\sqrt{2}) + \frac{\pi}{8}

, which matches option C.

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