Question-144975

Question Number 144975 by mim24 last updated on 01/Jul/21

Answered by ArielVyny last updated on 01/Jul/21

\int u^{'} \times u^{n} d u = \frac{1}{n + 1} u^{n + 1} + c t e

\int (1 + \frac{1}{x}) (x + l n (x)) d x = \frac{1}{2} {(x + l n x)}^{2}

Answered by som(math1967) last updated on 01/Jul/21

l e t (x + l n x) = z

(1 + \frac{1}{x}) d x = d z

\int z d z = \frac{z^{2}}{2} + C = \frac{{(x + l n x)}^{2}}{2} + C

Answered by imjagoll last updated on 01/Jul/21

let u = x + \ln x \Rightarrow du = (1 + \frac{1}{x}) dx

I = \int u du = \frac{u^{2}}{2} + c = \frac{{(x + \ln x)}^{2}}{2} + c

= \frac{{(\ln {xe}^{x})}^{2}}{2} + c

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