$\int \frac{1}{4-y^{2}} d y=-\frac{1}{4}(\ln |y-2|+\ln |y+2|)+C$ .

Name: $\int \frac{1}{4-y^{2}} d y=-\frac{1}{4}(\ln |y-2|+\ln |y+2|)+C$ .
Brand: Quizplus
Rating: 4.9 (37 reviews)

Question

Quizplus · Accepted Answer

The correct integral of $\frac{1}{4-y^{2}}$ is $\frac{1}{4}\ln\left|\frac{2-y}{2+y}\right|+C$, because $\frac{1}{4-y^{2}}$ can be rewritten as $\frac{1}{2^2-y^2}$, which leads to a partial fraction decomposition of $\frac{1}{2(2-y)} - \frac{1}{2(2+y)}$, and integrating these terms gives the correct result.

∫14−y2dy=−14(ln⁡∣y−2∣+ln⁡∣y+2∣)+C\int \frac{1}{4-y^{2}} d y=-\frac{1}{4}(\ln |y-2|+\ln |y+2|)+C∫4−y21​dy=−41​(ln∣y−2∣+ln∣y+2∣)+C

$\int \frac{1}{4-y^{2}} d y=-\frac{1}{4}(\ln |y-2|+\ln |y+2|)+C$