Question

$\int \log \sqrt{x+1}dx=$

• A. $\frac{1}{2}\left[\right(x+1\left)\log \right(x+1)-x]+c$
• B. $\frac{1}{2}\left[\right(x-1\left)\log \right(x+1)-x]+c$
• C. $\frac{1}{2}[x\log (x+1)-\frac{x^2}{2}]+c$
• D. $\frac{1}{2}\left[\right(x+1)\log (x+1)-\frac{x}{2}]+c$

Answer 1

The correct option is A $\frac{1}{2}\left[\right(x+1\left)\log \right(x+1)-x]+c$

Integrating by parts,

$\int 1log\sqrt{x+1}\cdot dx$

$=log\sqrt{x+1}\int 1dx-\int \frac{dlog\sqrt{x+1}}{dx}\cdot x\cdot dx.$

$x(log\sqrt{x+1}-\frac{1}{2}\int (\frac{1}{1+x.})\cdot xdx.$

$\frac{1}{2}log(x+1{)}^x-\frac{1}{2}[\int 1\cdot dx-log(1+x)]$

$=\frac{1}{2}xlog(x+1)-\frac{1}{2}\cdot x+\frac{1}{2}log(1+x)$

$=\frac{1}{2}\left[log\right(x+1\left)\right[x+1]-x]+c\cdot$