Question

$\int \frac{\left(x^{3}-1\right)}{\left(x^{4}+1\right)(x+1)} d x=$

• A. $\frac{1}{4} \ln \left(1+x^{4}\right)+\frac{1}{3} \ln \left(1+x^{3}\right)+c$
• B. $\frac{1}{4} \ln \left(1+x^{4}\right)-\frac{1}{3} \ln \left(1+x^{3}\right)+c$
• C. $\frac{1}{4} \ln \left(1+x^{4}\right)-\ln (1+x)+c$
• D. $\frac{1}{4} \ln \left(1+x^{4}\right)+\ln (1+x)+c$

Answer 1

Answer: (C)

$\int \frac{x^{3}-1}{\left(x^{4}+1\right)(x+1)} d x =\int \frac{\left(x^{4}+x^{3}\right)-\left(x^{4}+1\right)}{\left(x^{4}+1\right)(x+1)} d x$
$=\int \frac{x^{3}}{x^{4}+1} d x-\int \frac{1}{x+1} d x$
$=\frac{1}{4} \ln \left(x^{4}+1\right)-\ln (x+1)+c$