If I - -dx-ex - 23- then I equals

The correct option is A 1/8 x 1/8log (e^x + 2) + e^x + 3/4 (e^x + 2)^2 + C I=∫ #160; dx /( e^ x +2 ) ^ 3 Substitute e ^ x +2=t I=∫ dt / ...

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Standard XII

Mathematics

Special Integrals - 1

If I = ∫dx/...

Question

If $I = \int \frac{d x}{(e^{x} + 2)^{3}}$ , then I equals

\frac{1}{8} x - \frac{1}{8} log (e^{x} + 2) + \frac{e^{x} + 3}{4 (e^{x} + 2)^{2}} + C

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\frac{1}{8} x + \frac{1}{8} log (e^{x} + 2) + \frac{e^{x}}{4 (e^{x} + 2)^{2}} + C

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\frac{1}{8} x + \frac{1}{8} log (e^{x} + 2) + \frac{e^{x}}{(e^{x} + 2)^{2}} + C

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none of these

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Solution

The correct option is A $\frac{1}{8} x - \frac{1}{8} log (e^{x} + 2) + \frac{e^{x} + 3}{4 (e^{x} + 2)^{2}} + C$
$I = \int \frac{d x}{{(e^{x} + 2)}^{3}}$
Substitute $e^{x} + 2 = t$
$I = \int \frac{d t}{t^{3} (t - 2)}$
Now $\frac{1}{t^{3} (t - 2)} = \frac{1}{8 (t - 2)} - \frac{1}{8 t} - \frac{1}{4 t^{2}} - \frac{1}{2 t^{3}}$

$∴ I = \int (\frac{1}{8 (t - 2)} - \frac{1}{8 t} - \frac{1}{4 t^{2}} - \frac{1}{2 t^{3}}) d t$

$= \int \frac{1}{8 (t - 2)} d t - \int \frac{1}{8 t} d t - \int \frac{1}{4 t^{2}} d t - \int \frac{1}{2 t^{3}} d t$

$= \frac{1}{8} log | t - 2 | - \frac{1}{8} log t + \frac{1}{4 t} + \frac{1}{4 t^{2}} + c$

$= \frac{1}{8} log | e^{x} + 2 - 2 | - \frac{1}{8} log | e^{x} + 2 | + \frac{1}{4 (e^{x} + 2)} + \frac{1}{4 {(e^{x} + 2)}^{2}} + c$

$= \frac{1}{8} x - \frac{1}{8} log | e^{x} + 2 | + \frac{e^{x} + 3}{4 {(e^{x} + 2)}^{2}} + c$

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If I=∫dx(ex+2)3, then I equals

If $I = \int \frac{d x}{(e^{x} + 2)^{3}}$ , then I equals