The value of the integral _2^4 x/(x^2+1) dx is : Option 1: 1/2(5/17) Option 2: 1/2(17/5) Option 3: 2(5/17) Option 4: 2(17/5)

Answer: 1/2(17/5). Solution: Let u = x^2+1, du = 2x dx. x/(x^2+1)dx = 1/2(x^2+1). Evaluating from 2 to 4: 1/2[ 17 - 5] = 1/2(17/5).

Calculus

Integrals

Easy

common

The value of the integral $\int_{2}^{4} \frac{x}{x^2+1} dx$ is :

$\frac{1}{2}\log\left(\frac{5}{17}\right)$

$\frac{1}{2}\log\left(\frac{17}{5}\right)$

$2\log\left(\frac{5}{17}\right)$

$2\log\left(\frac{17}{5}\right)$

✅ Correct Option: 2

Let $u = x^2+1$ , $du = 2x\,dx$ .

$\int \frac{x}{x^2+1}dx = \frac{1}{2}\log(x^2+1)$ .

Evaluating from 2 to 4: $\frac{1}{2}[\log 17 - \log 5] = \frac{1}{2}\log\left(\frac{17}{5}\right)$ .

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Choose the correct answer from the options given below:

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