Integrated the function: e ^ 2 x - e ^ - 2 x e ^ 2 x + e ^ - 2 x d x.

Let I = e ^ 2 x - e ^ - 2 x e ^ 2 x + e ^ - 2 x d x Put e^ 2x + e^ -2x = t ( 2 e ^ 2 x - 2 e ^ - 2 x ) d x = d t [ d d x ( e ^ a x ) = a e ^ a x ] 2 ( e ^ 2 x - e ^ - 2 x ) d x = d t ( e ^ 2 x - e ^ - 2 x ) d x = d t 2 I = e ^ 2 x - e ^ - 2 x e ^ 2 x + e ^ - 2 x d x = 1 2 d t t = 1 2 | t | + C = 1 2 | e ^ 2 x + e ^ - 2 x | + C [ put t = e ^ 2 x + e ^ - 2 x ] I = 1 2 | e ^ 2 x + e ^ - 2 x | + C

Integrated the function: e ^ 2 x - e ^ - 2 x e ^ 2 x + e ^ - 2 x …

Download App

Question

Integrated the function: $\int \frac { e ^ { 2 x } - e ^ { - 2 x } } { e ^ { 2 x } + e ^ { - 2 x } } d x.$

✓

Answer

Let $I = \int \frac { e ^ { 2 x } - e ^ { - 2 x } } { e ^ { 2 x } + e ^ { - 2 x } } d x$
Put $e^{2x} + e^{-2x} = t$
$\Rightarrow \left( 2 e ^ { 2 x } - 2 e ^ { - 2 x } \right) d x = d t$ $\left[ \because \frac { d } { d x } \left( e ^ { a x } \right) = a e ^ { a x } \right]$
$\Rightarrow 2\left( e ^ { 2 x } - e ^ { - 2 x } \right) d x = { d t } $
$\Rightarrow \left( e ^ { 2 x } - e ^ { - 2 x } \right) d x = \frac { d t } { 2 }$
$\therefore I = \int \frac { e ^ { 2 x } - e ^ { - 2 x } } { e ^ { 2 x } + e ^ { - 2 x } } d x$
$ = \frac { 1 } { 2 } \int \frac { d t } { t } $
$= \frac { 1 } { 2 } \log | t | + C$
$= \frac { 1 } { 2 } \log \left| e ^ { 2 x } + e ^ { - 2 x } \right| + C \quad \text { [ put } t = e ^ { 2 x } + e ^ { - 2 x } ]$
$\therefore I =\frac { 1 } { 2 } \log \left| e ^ { 2 x } + e ^ { - 2 x } \right| + C$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "3 Marks" from Integrals→Integrals — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Integrals — Maths STD 12 Science — Question

Answer

Need a full question paper?

Explore more

Similar questions