e-4x-1-dx-

Question Number 94520 by john santu last updated on 19/May/20

\int \sqrt{e^{4 x} + 1} d x

Answered by niroj last updated on 19/May/20

\int \sqrt{{(e^{x})}^{4} + 1} dx

Put e^{x} = t

e^{x} . dx = dt

dx = \frac{1}{t} dt

= \int \sqrt{t^{4} + 1} . \frac{1}{t} dt

= \int \sqrt{t^{2} (t^{2} + \frac{1}{t^{2}})} \frac{1}{t} dt

= \int t \sqrt{t^{2} + \frac{1}{t^{2}}} \frac{1}{t} dt

= \int \sqrt{{(t + \frac{1}{t})}^{2} - 2} dt

= \int \sqrt{{(t + \frac{1}{t})}^{2} - {(\sqrt{2})}^{2}} dt

= \frac{(t + \frac{1}{t}) \sqrt{t^{2} + \frac{1}{t^{2}}}}{2} - \frac{{(\sqrt{2})}^{2}}{2} \log (t + \frac{1}{t} + \sqrt{t^{2} + \frac{1}{t^{2}}}) + C

= \frac{\frac{t^{2} + 1}{t} . \frac{\sqrt{t^{4} + 1}}{t}}{2} - \log (\frac{t^{2} + 1}{t} + \frac{\sqrt{t^{4} + 1}}{t}) + C

= \frac{(t^{2} + 1) \sqrt{t^{4} + 1}}{{2 t}^{2}} - \log (\frac{t^{2} + 1 + \sqrt{t^{4} + 1}}{t}) + C

= \frac{(e^{2 x} + 1) (\sqrt{e^{4 x} + 1})}{{2 e}^{2 x}} - \log (\frac{e^{2 x} + 1 + \sqrt{e^{4 x} + 1}}{e^{x}}) + C / / .

Commented by i jagooll last updated on 19/May/20

cooll man ��

Commented by peter frank last updated on 19/May/20

t h a n k y o u

Commented by Mr.D.N. last updated on 19/May/20

great��

Commented by niroj last updated on 19/May/20

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Answered by john santu last updated on 19/May/20

let u = e^{4 x} \Rightarrow d x = \frac{d u}{4 u}

\int \frac{\sqrt{u + 1}}{4 u} d u = \frac{1}{4} \int \frac{\sqrt{u + 1}}{u} d u

s e t v = \sqrt{u + 1} \Rightarrow d u = 2 v d v

\int \frac{1}{4} \frac{2 v^{2}}{v^{2} - 1} d v = \frac{1}{2} \int (1 + \frac{1}{v^{2} - 1}) d v

= \frac{1}{2} (v + \frac{1}{2} \ln ∣ \frac{v - 1}{v + 1} ∣) + c

= \frac{1}{2} \sqrt{e^{4 x} + 1} + \frac{1}{4} \ln ∣ \frac{\sqrt{e^{4 x} + 1} - 1}{\sqrt{e^{4 x} + 1} + 1} ∣ + c

Commented by i jagooll last updated on 19/May/20

waw \dots . funtastic

Answered by MJS last updated on 19/May/20

why not all at once ?

\int \sqrt{e^{4 x} + 1} d x =

[t = \sqrt{e^{4 x} + 1} \to d x = \frac{\sqrt{e^{4 x} + 1}}{{2 e}^{4 x}} d t]

= \frac{1}{2} \int \frac{t^{2}}{t^{2} - 1} d t = \frac{t}{2} + \frac{1}{4} \ln \frac{t - 1}{t + 1} = \dots

Answered by mathmax by abdo last updated on 19/May/20

I = \int \sqrt{e^{4 x} + 1} dx we do the changement \sqrt{e^{4 x} + 1} = t \Rightarrow

e^{4 x} = t^{2} - 1 \Rightarrow 4 x = \ln (t^{2} - 1) \Rightarrow x = \frac{1}{4} \ln (t^{2} - 1) \Rightarrow dx = \frac{t}{2 (t^{2} - 1)} dt

\Rightarrow I = \frac{1}{2} \int t^{2} \times \frac{dt}{t^{2} - 1} = \frac{1}{2} \int \frac{t^{2} - 1 + 1}{t^{2} - 1} dt

= \frac{t}{2} + \frac{1}{4} \int (\frac{1}{t - 1} - \frac{1}{t + 1}) dt = \frac{t}{2} + \frac{1}{4} \ln ∣ \frac{t - 1}{t + 1} ∣ + C

= \frac{1}{2} \sqrt{e^{4 x} + 1} + \frac{1}{4} \ln ∣ \frac{\sqrt{e^{4 x} + 1} - 1}{\sqrt{e^{4 x} + 1} + 1} ∣ + C