Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDefinite Integration2 Marks
MCQ
$\int_1^e \frac{1+\log x}{x} d x=$
✓
$\frac{3}{2}$
B
$\frac{1}{7}$
C
$\frac{1}{e}$
D
$\frac{2}{3}$
✓
Answer
Correct option: A.
$\frac{3}{2}$
(A) Put $1+\log x= t \Rightarrow \frac{1}{x} d x= dt$ When $x=1, t =1$ and when $x= e , t =2$ $\therefore \quad \int_1^{ e } \frac{1+\log x}{x} d x=\int_1^2 tdt =\left[\frac{ t ^2}{2}\right]_1^2=\frac{3}{2}$
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