Question

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

• A. $\frac{1}{8}x-\frac{1}{8}\log (e^x+2)+\frac{e^x+3}{4(e^x+2{)}^2}+C$
• B. $\frac{1}{8}x+\frac{1}{8}\log (e^x+2)+\frac{e^x}{4(e^x+2{)}^2}+C$
• C. $\frac{1}{8}x+\frac{1}{8}\log (e^x+2)+\frac{e^x}{(e^x+2{)}^2}+C$
• D. none of these

Answer 1

The correct option is A $\frac{1}{8}x-\frac{1}{8}\log (e^x+2)+\frac{e^x+3}{4(e^x+2{)}^2}+C$

$I=\int \frac{dx}{{(e^x+2)}^3}$

Substitute $e^x+2=t$

$I=\int \frac{dt}{t^3(t-2)}$

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

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

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

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

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

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