The value of int_0^1 frac{x^4 + 1}{x^2 + 1} , | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $I = \int_0^1 \frac{x^4 + 1}{x^2 + 1} \, dx$. We can rewrite the integrand as $\frac{x^4 - 1 + 2}{x^2 + 1} = \frac{(x^2 - 1)(x^2 + 1) + 2}{x^2

Vedclass · Accepted Answer

$\frac{1}{6}(3\pi - 4)$

Vedclass · Answer

(A) Let $I = \int_0^1 \frac{x^4 + 1}{x^2 + 1} \, dx$. We can rewrite the integrand as $\frac{x^4 - 1 + 2}{x^2 + 1} = \frac{(x^2 - 1)(x^2 + 1) + 2}{x^2 + 1} = x^2 - 1 + \frac{2}{x^2 + 1}$. Now,integrate term by term: $I = \int_0^1 (x^2 - 1) \, dx + 2 \int_0^1 \frac{1}{x^2 + 1} \, dx$. Evaluating the integrals: $I = \left[ \frac{x^3}{3} - x ight]_0^1 + 2 \left[ 	an^{-1} x ight]_0^1$. $I = \left( \frac{1}{3} - 1 ight) - (0) + 2 \left( 	an^{-1}(1) - 	an^{-1}(0) ight)$. $I = -\frac{2}{3} + 2 \left( \frac{\pi}{4} - 0 ight)$. $I = -\frac{2}{3} + \frac{\pi}{2} = \frac{3\pi - 4}{6}$.

The value of $\int_0^1 \frac{x^4 + 1}{x^2 + 1} \, dx$ is

Similar Questions

The value$(s)$ of $\int_0^1 \frac{x^4(1-x)^4}{1+x^2} d x$ is (are)

The limit of the area under the curve $y = e^{-x}$ from $x = 0$ to $x = h$ as $h \rightarrow \infty$ is:

$\int_{2}^{3} \frac{dx}{x^{2}+x} = $

$\int_1^{\sqrt{3}} \frac{1}{1 + x^2} dx$ is equal to

Evaluate the definite integral: $\int_0^{\pi /8} \frac{\sec^2 2x}{2} \, dx$

Vedclass Test Series

Exam Paper Generator

Online Exam Module