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$\int\Large{\frac{x+3}{(x+4)^2}}\normalsize e^xdx$=___________.

1 Answer

Toolbox:
  • $\int e^x[f(x)+f'(x)]dx=e^xf(x)+c.$
  • $\large\frac{d}{dx}\big(\frac{1}{x}\big)=\large\frac{-1}{x^2}$
Step 1:
Let $I=\int\large\frac{x+3}{(x+4)^2}$$e^xdx.$
Consider $\large\frac{x+3}{(x+4)^2}$
Add and subtract 1 to the numerator,
$\large\frac{x+3+1-1}{(x+4)^2}=\large\frac{(x+4)-1}{(x+4)^2}$
On splitting the terms we get,
$\large\frac{x+4}{(x+4)^2}-\frac{1}{(x+4)^2}=\large\frac{1}{(x+4)}-\frac{1}{(x+4)^2}$
 
Step 2:
If $f(x)=\large\frac{1}{x+4}$,then $f'(x)=\large\frac{-1}{(x+4)^2}$
Clearly the integrand is of the form
$\int e^x[f(x)+f'(x)]dx=e^xf(x)+c.$
$I=\int e^x\begin{bmatrix}\large\frac{1}{(x+4)}-\frac{1}{(x+4)^2}\end{bmatrix}$$dx$
$\;\;=e^x\big(\large\frac{1}{x+4}\big)$$+c$
answered Apr 24, 2013 by sreemathi.v
 
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