$\int \frac{7 x^{2}+6}{x^{2}+1} d x=7 x-\arctan x+C$ .

Name: $\int \frac{7 x^{2}+6}{x^{2}+1} d x=7 x-\arctan x+C$ .
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Question

Quizplus · Accepted Answer

The integral of $\frac{7 x^{2}+6}{x^{2}+1}$ is indeed $7 x-\arctan x+C$. This can be verified by using the method of partial fractions or by using a trigonometric substitution.

∫7x2+6x2+1dx=7x−arctan⁡x+C\int \frac{7 x^{2}+6}{x^{2}+1} d x=7 x-\arctan x+C∫x2+17x2+6​dx=7x−arctanx+C

$\int \frac{7 x^{2}+6}{x^{2}+1} d x=7 x-\arctan x+C$