Question Bank [2022] — Maths (commerce) STD 12 Commerce / Arts — Question
Maharashtra BoardEnglish MediumSTD 12 Commerce / ArtsMaths (commerce)Question Bank [2022]3 Marks
Question
Evaluate $\int \frac{1}{\sqrt{3 x^2+8}} d x$
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Answer
$\text { Let } I =\int \frac{1}{\sqrt{3 x^2+8}} d x$
$=\frac{1}{\sqrt{3}} \int \frac{1}{\sqrt{x^2+\frac{8}{3}}} d x$
$=\frac{1}{\sqrt{3}} \int \frac{1}{\sqrt{x^2+\left(\frac{2 \sqrt{2}}{\sqrt{3}}\right)^2}} d x$
$=\frac{1}{\sqrt{3}} \log \left|x+\sqrt{x^2+\left(\frac{2 \sqrt{2}}{\sqrt{3}}\right)^2}\right|+ c _1$
$=\frac{1}{\sqrt{3}} \log \left|x+\sqrt{x^2+\frac{8}{3}}\right|+ c _1$
$=\frac{1}{\sqrt{3}} \log \left|\frac{\sqrt{3} x+\sqrt{3 x^2+8}}{\sqrt{3}}\right|+ c _1$
$=\frac{1}{\sqrt{3}} \log \left|\sqrt{3} x+\sqrt{3 x^2+8}\right|-\frac{1}{\sqrt{3}} \log \sqrt{3}+ c _1$
$\therefore I =\frac{1}{\sqrt{3}} \log \left|\sqrt{3} x+\sqrt{3 x^2+8}\right|+ c ,$
$\text { where } c = c _1-\frac{1}{\sqrt{3}} \log \sqrt{3}$
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