I- -1-e2x - e-2x dx - A 842 tan -1 e 2x - C then A is equal to

I=∫ 1 e ^ 2x + e ^ 2x dx Put t= e ^ x ⇒ dt= e ^ x dx I=∫ t t ^ 4 +1 dt Put u= t ^ 2 ⇒ du=2tdt I= 1 2 ∫ 1 u ^ 2 +1 du = 1 2 ta ...

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Byju's Answer

Standard XII

Mathematics

Parametric Differentiation

I= ∫1/e2x + e...

Question

$I = \int \frac{1}{e^{2 x} + e^{- 2 x}} d x = \frac{A}{842} {tan}^{- 1} e^{2 x} + C$ then A is equal to

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Solution

$I = \int \frac{1}{e^{2 x} + e^{- 2 x}} d x$
Put $t = e^{x} \Rightarrow d t = e^{x} d x$
$I = \int \frac{t}{t^{4} + 1} d t$
Put $u = t^{2} \Rightarrow d u = 2 t d t$
$I = \frac{1}{2} \int \frac{1}{u^{2} + 1} d u = \frac{1}{2} {tan}^{- 1} u + c = \frac{1}{2} {tan}^{- 1} t^{2} + c = \frac{1}{2} {tan}^{- 1} e^{2 x} + c$
Therefore $A = 421$

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I=∫1e2x+e−2xdx=A842tan−1e2x+C then A is equal to

I=∫1e2x+e−2xdxPut t=ex⇒dt=exdxI=∫tt4+1dtPut u=t2⇒du=2tdtI=12∫1u2+1du=12tan−1u+c=12tan−1t2+c=12tan−1e2x+cTherefore A=421

$I = \int \frac{1}{e^{2 x} + e^{- 2 x}} d x = \frac{A}{842} {tan}^{- 1} e^{2 x} + C$ then A is equal to