Question

If $A={\int }_0^1\frac{e^t}{1+t}dt$ then ${\int }_0^1{e}^{t}ln(1+t)dt=$

• A. $eln2-A$
• B. $eln2+A$
• C. $Aeln2$
• D. $Aln2$

Answer 1

The correct option is A $eln2-A$

$A=\underset{0}{\overset{1}{\int }}\frac{e^t}{1+t}dt$

Now,

$\underset{0}{\overset{1}{\int }}{e}^{t}ln(1+t)dt$

Integrating by parts taking $ln(1+t)$ as the first function.

$={\left[ln(1+t)e^t\right]}_0^1-{\int }_0^1\frac{1}{1+t}\times e^{t}dt$

$=e\ln 2-A$