find ∫e^x lnx dx


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Question Number 40587 by mondodotto@gmail.com last updated on 24/Jul/18

$find \int e^{x} \ln x d x$
Commented by prof Abdo imad last updated on 25/Jul/18
$∫ e^x ln(x)dx = ∫(Σ_(n=0) ^∞ (x^n /(n!)))ln(x)dx =Σ_(n=0) ^∞ (1/(n!)) ∫ x^n ln(x)dx by parts A_n =∫ x^n ln(x)dx=(1/(n+1))x^(n+1) ln(x)−∫ (1/(n+1))x^n dx =(1/(n+1)) x^(n+1) ln(x) −(1/((n+1)^2 )) +λ⇒ ∫ e^x ln(x)dx=Σ_(n=0) ^∞ (1/(n!)){ (1/(n+1)) x^(n+1) ln(x)−(1/((n+1)^2 )) +λ} =ln(x)Σ_(n=0) ^∞ (x^(n+1) /((n+1)n!)) −Σ_(n=0) ^∞ (1/((n+1)^2 n!)) +λe let w(x)=Σ_(n=0) ^∞ (x^(n+1) /((n+1)n!)) we have w^′ (x) =Σ_(n=0) ^∞ (x^n /(n!)) =e^x ⇒w(x)=e^x +c w(0)=0=1+c ⇒c=−1 ⇒w(x)=e^x −1 ⇒ ∫ e^x ln(x)dx =(e^x −1)ln(x)−Σ_(n=0) ^∞ (1/((n+1)^2 n!)) +λe ...be continued...$
$\int e^{x} l n (x) d x = \int (\sum_{n = 0}^{\infty} \frac{x^{n}}{n!}) l n (x) d x$ $= \sum_{n = 0}^{\infty} \frac{1}{n!} \int x^{n} l n (x) d x b y p a r t s$ $A_{n} = \int x^{n} l n (x) d x = \frac{1}{n + 1} x^{n + 1} l n (x) - \int \frac{1}{n + 1} x^{n} d x$ $= \frac{1}{n + 1} x^{n + 1} l n (x) - \frac{1}{{(n + 1)}^{2}} + λ \Rightarrow$ $\int e^{x} l n (x) d x = \sum_{n = 0}^{\infty} \frac{1}{n!} {\frac{1}{n + 1} x^{n + 1} l n (x) - \frac{1}{{(n + 1)}^{2}} + λ}$ $= l n (x) \sum_{n = 0}^{\infty} \frac{x^{n + 1}}{(n + 1) n!} - \sum_{n = 0}^{\infty} \frac{1}{{(n + 1)}^{2} n!} + λ e$ $l e t w (x) = \sum_{n = 0}^{\infty} \frac{x^{n + 1}}{(n + 1) n!} w e h a v e$ $w^{^{'}} (x) = \sum_{n = 0}^{\infty} \frac{x^{n}}{n!} = e^{x} \Rightarrow w (x) = e^{x} + c$ $w (0) = 0 = 1 + c \Rightarrow c = - 1 \Rightarrow w (x) = e^{x} - 1 \Rightarrow$ $\int e^{x} l n (x) d x = (e^{x} - 1) l n (x) - \sum_{n = 0}^{\infty} \frac{1}{{(n + 1)}^{2} n!}$ $+ λ e . . . b e c o n t i n u e d . . .$
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