How do I evaluate $\int \frac{d x}{e^{x} (3 e^{x} + 2)}$ $intdx/(e^(x)(3e^(x)+2))$ ?

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How do I evaluate $\int \frac{d x}{e^{x} (3 e^{x} + 2)}$ $intdx/(e^(x)(3e^(x)+2))$ ?

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Answer 1 · 2018-06-18T15:55:49

The answer is $= - \frac{1}{2 e^{x}} - \frac{3}{4} x + \frac{3}{4} ln (3 e^{x} + 2) + C$ $=-1/(2e^x)-3/4x+3/4ln(3e^x+2)+C$

Explanation:

Let $e^{x} = u$ $e^x=u$ , $\Rightarrow$ $=>$ , $d u = e^{x} d x$ $du=e^xdx$

Therefore, the integral is

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

$= \int \frac{d u}{u^{2} (3 u + 2)}$ $=int(du)/(u^2(3u+2))$

Perform the decomposition into partial fractions

$\frac{1}{u^{2} (3 u + 2)} = \frac{A}{u^{2}} + \frac{B}{u} + \frac{C}{3 u + 2}$ $1/(u^2(3u+2))=A/u^2+B/u+C/(3u+2)$

$= \frac{A (3 u + 2) + B u (3 u + 2) + C u^{2}}{u^{2} (3 u + 2)}$ $=(A(3u+2)+Bu(3u+2)+Cu^2)/(u^2(3u+2))$

The denominator is the same, compare the numerators

$1 = A (3 u + 2) + B u (3 u + 2) + C u^{2}$ $1=A(3u+2)+Bu(3u+2)+Cu^2$

Let $u = 0$ $u=0$ , $\Rightarrow$ $=>$ , $1 = 2 A$ $1=2A$ , $\Rightarrow$ $=>$ , $A = \frac{1}{2}$ $A=1/2$

Coefficients of $u$ $u$

$0 = 3 A + 2 B$ $0=3A+2B$ , $\Rightarrow$ $=>$ , $B = - \frac{3}{2} A = - \frac{3}{4}$ $B=-3/2A=-3/4$

Coefficients of $u^{2}$ $u^2$

$0 = 3 B + C$ $0=3B+C$ , $\Rightarrow$ $=>$ , $C = - 3 B = \frac{9}{4}$ $C=-3B=9/4$

Therefore,

$\frac{1}{u^{2} (3 u + 2)} = \frac{\frac{1}{2}}{u^{2}} + \frac{- \frac{3}{4}}{u} + \frac{\frac{9}{4}}{3 u + 2}$ $1/(u^2(3u+2))=(1/2)/u^2+(-3/4)/u+(9/4)/(3u+2)$

So,

$\int \frac{d u}{u^{2} (3 u + 2)} = \frac{1}{2} \int \frac{d u}{u^{2}} - \frac{3}{4} \int \frac{d u}{u} + \frac{9}{4} \int \frac{d u}{3 u + 2}$ $int(du)/(u^2(3u+2))=1/2int(du)/u^2-3/4int(du)/u+9/4int(du)/(3u+2)$

$= - \frac{1}{2 u} - \frac{3}{4} ln u + \frac{3}{4} ln (3 u + 2)$ $=-1/(2u)-3/4lnu+3/4ln(3u+2)$

$= - \frac{1}{2 e^{x}} - \frac{3}{4} ln (e^{x}) + \frac{3}{4} ln (3 e^{x} + 2) + C$ $=-1/(2e^x)-3/4ln(e^x)+3/4ln(3e^x+2)+C$

$= - \frac{1}{2 e^{x}} - \frac{3}{4} x + \frac{3}{4} ln (3 e^{x} + 2) + C$ $=-1/(2e^x)-3/4x+3/4ln(3e^x+2)+C$

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How do I evaluate $\int \frac{d x}{e^{x} (3 e^{x} + 2)}$ $intdx/(e^(x)(3e^(x)+2))$ ?

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