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Question Bank [2022] — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsQuestion Bank [2022]4 Marks
Question
$\int \sqrt{\frac{ e ^{3 x}- e ^{2 x}}{ e ^x+1}} d x$
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Answer

$ \text { Let } I =\int \sqrt{\frac{ e ^{3 x}- e ^{2 x}}{ e ^x+1}} d x$
$=\int \sqrt{\frac{ e ^{2 x}\left( e ^x-1\right)}{ e ^x+1}} d x$
$=\int e ^x \sqrt{\frac{ e ^x-1}{ e ^x+1}} d x $
Put $e^x=t$
$ \therefore e^{ x } d x = dt$
$\therefore I =\int \sqrt{\frac{ t -1}{ t +1}} dt$
$=\int \sqrt{\frac{ t -1}{ t +1} \times \frac{ t -1}{ t -1} dt }$
$=\int \frac{ t -1}{\sqrt{ t ^2-1}} dt$
$=\int\left(\frac{ t }{\sqrt{ t ^2-1}}-\frac{1}{\sqrt{ t ^2-1}}\right) dt$
$=\int \frac{ t }{\sqrt{ t ^2-1}} dt -\int \frac{1}{\sqrt{ t ^2-1}} dt $
$ =I_1-I_2 \quad \ldots \ldots . .(i)$
$I_1=\int \frac{t}{\sqrt{t^2-1}} d t $
Put $t^2-1=a$
$ \therefore 2 t d t = da$
$\therefore I _1=\frac{1}{2} \int \frac{ da }{\sqrt{ a }}$
$=\frac{1}{2} \int a ^{\frac{1}{2}} da$
$=\frac{1}{2}\left(\frac{ a ^{\frac{1}{2}}}{\frac{1}{2}}\right)+ c _1$
$=\sqrt{ a }+ c _1$
$=\sqrt{ t ^2-1}+ c _1$
$\therefore I _1=\sqrt{ e ^{2 x}-1}+ c _1 \quad \ldots \ldots . \text { (ii) }$
$I _2=\int \frac{1}{\sqrt{ t ^2-1^2}} dt$
$=\log \left| t +\sqrt{ t ^2-1^2}\right|+ c _2$
$=\log \left| e ^x+\sqrt{ e ^{2 x}-1}\right|+ c _2\ldots(iii) $
From (i), (ii) and (iii), we get
$\left|=\sqrt{ e ^{2 x}-1}-\log \right| e ^x+\sqrt{ e ^{2 x}-1} \mid+ c _t$
where $c = c _1- c _2$

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