∀x>0, ∫ex(x−2)dxx(x2+ex) is equal to

I=∫xex−2exx3+xex=∫3x2+xex+ex−3x2−3exx3+xexdx∫3x2+xex+exx3+xexdx−3∫1xdx3x2+xex+exdx=dt=∫dtt−3ln|x|+C⇒ln|t|−3ln|x|+C⇒ln(x3+xexx3)+C=ln(1+exx2)+C

∀x>0, ∫ex(x−2)dxx(x2+ex) is equal to

I=∫xex−2exx3+xex=∫3x2+xex+ex−3x2−3exx3+xexdx∫3x2+xex+exx3+xexdx−3∫1xdx3x2+xex+exdx=dt=∫dtt−3ln|x|+C⇒ln|t|−3ln|x|+C⇒ln(x3+xexx3)+C=ln(1+exx2)+C

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$\forall x > 0, \int \frac{e^{x} (x - 2) d x}{x (x^{2} + e^{x})}$ is equal to

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$l n (1 + \frac{e^{x}}{x^{2}}) + C$

$l n (\frac{- 1}{2} + \frac{e^{x}}{x^{2}}) + C$

$l n (2 + \frac{e^{x}}{x^{2}}) + C$

$l n (x + \frac{e^{x}}{x^{2}}) + C$

answer is A.

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Detailed Solution

$\begin{array}{l} I = \int \frac{x e^{x} - 2 e^{x}}{x^{3} + x e^{x}} \\ = \int \frac{3 x^{2} + x e^{x} + e^{x} - 3 x^{2} - 3 e^{x}}{x^{3} + x e^{x}} d x \end{array}$

$\int \frac{3 x^{2} + x e^{x} + e^{x}}{x^{3} + x e^{x}} d x - 3 \int \frac{1}{x} d x$

$\begin{array}{l} 3 x^{2} + x e^{x} + e^{x} d x = d t \\ = \int \frac{d t}{t} - 3 l n | x | + C \\ \Rightarrow l n | t | - 3 l n | x | + C \Rightarrow l n (\frac{x^{3} + x e^{x}}{x^{3}}) + C \\ = l n (1 + \frac{e^{x}}{x^{2}}) + C \end{array}$