Maharashtra BoardEnglish MediumSTD 12 Commerce / ArtsMaths (commerce)Integration (p-1)2 Marks
Question
Evalute : $\int x \sqrt{1+x^2} d x$
✓
Answer
$ \text { Let } I=\int x \sqrt{1+x^2} d x=\int \sqrt{1+x^2} \cdot x d x $ Put $1+x^2=t$ $ \begin{aligned} & \therefore 2 x d x=d t \quad \therefore x d x=\frac{d t}{2} \\ & \begin{aligned} \therefore I & =\int \sqrt{t} \frac{d t}{2}=\frac{1}{2} \int t^{\frac{1}{2}} d t \\ & =\frac{1}{2} \cdot \frac{t^{\frac{3}{2}}}{(3 / 2)}+c \\ & =\frac{1}{3}\left(1+x^2\right)^{\frac{3}{2}}+c . \end{aligned} \end{aligned} $
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