Evaluate the integral. $\int_{0}^{1} \frac{x}{x+8} d x$
A) $1-9 \ln 8+\sqrt{8} \ln 16$
B) $\frac{1}{8 \ln 9}-\ln \sqrt{8}$
C) $1-8(\ln 9-\ln 8)$
D) $9-8 \ln 9+\ln \sqrt{8}$
E) $\text { None of these }$

Question

Quizplus · Accepted Answer

Use u-substitution with $u=x+8$ and $du=dx$. The integral becomes 
\begin{align*}
\int_8^9 \frac{u-8}{u}du &= \int_8^9 1-\frac{8}{u} \, du \
&= \left[u-8\ln |u| ight]_{8}^9 \
&= 1-8(\ln 9 - \ln 8) \
&= \boxed{	extbf{(C)}}.
\end{align*}

Evaluate the Integral ∫01xx+8dx\int_{0}^{1} \frac{x}{x+8} d x∫01​x+8x​dx A) 1−9ln⁡8+8ln⁡161-9 \ln 8+\sqrt{8} \ln 161−9ln8+8​ln16

Evaluate the Integral $\int_{0}^{1} \frac{x}{x+8} d x$ A) $1-9 \ln 8+\sqrt{8} \ln 16$