Evaluate -ex-e2x-6ex-5dx-The ans is -1-4log - ex-1-ex- k -Cwhat is k

I=∫e^x/e^2x+6e^x+5dx=∫e^x/ ( e^x+6e^x+5 )dx Let e^x=t or e^x dx=dt ∴ I=∫dt/t^2+6t+5 =∫1/ ( t+3 )^2 2^2dt =1/2× 2log | t+3 2/t+3 ...

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Byju's Answer

Standard XII

Mathematics

Special Integrals - 1

Evaluate ∫e...

Question

Evaluate $\int \frac{e^{x}}{e^{2 x} + 6 e^{x} + 5} d x .$
The ans is $= \frac{1}{4} l o g ∣ ∣ ∣ \frac{e^{x} + 1}{e^{x} + k} ∣ ∣ ∣ + C$
what is k

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Solution

$I = \int \frac{e^{x}}{e^{2 x} + 6 e^{x} + 5} d x = \int \frac{e^{x}}{(e^{x} + 6 e^{x} + 5)} d x$
Let $e^{x} = t$ or $e^{x} d x = d t$
$∴ I = \int \frac{d t}{t^{2} + 6 t + 5}$
$= \int \frac{1}{{(t + 3)}^{2} - 2^{2}} d t$
$= \frac{1}{2 \times 2} l o g ∣ ∣ ∣ \frac{t + 3 - 2}{t + 3 + 2} ∣ ∣ ∣ + C = \frac{1}{4} l o g ∣ ∣ ∣ \frac{e^{x} + 1}{e^{x} + 5} ∣ ∣ ∣ + C$

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Find

k

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Evaluate ∫exe2x+6ex+5dx.The ans is =14log∣∣∣ex+1ex+k∣∣∣+Cwhat is k

I=∫exe2x+6ex+5dx=∫ex(ex+6ex+5)dxLet ex=t or exdx=dt∴I=∫dtt2+6t+5=∫1(t+3)2−22dt=12×2log∣∣∣t+3−2t+3+2∣∣∣+C=14log∣∣∣ex+1ex+5∣∣∣+C

Evaluate $\int \frac{e^{x}}{e^{2 x} + 6 e^{x} + 5} d x .$
The ans is $= \frac{1}{4} l o g ∣ ∣ ∣ \frac{e^{x} + 1}{e^{x} + k} ∣ ∣ ∣ + C$
what is k