eiraszero11cu

2021-12-26

Evaluate the indefinite integral.
$\int \frac{x+\mathrm{ln}x}{x}dx$

Corgnatiui

Step 1
The indefinite integral is given $\int \frac{x+\mathrm{ln}x}{x}dx$
Simplify the indefinite integral.
$\int \frac{x+\mathrm{ln}x}{x}dx=\int \frac{x}{x}dx+\int \frac{\mathrm{ln}x}{x}dx$
$=\int 1\cdot dx+\int \frac{\mathrm{ln}x}{x}dx$
$=x+\int \frac{\mathrm{ln}x}{x}dx$
Substitute $\mathrm{ln}x⇒$.
$\frac{1}{x}dx⇒dt$
Step 2
Apply the values integral,
$\int \frac{\mathrm{ln}x}{x}dx=\int tdt$
$=\frac{{t}^{2}}{2}+c$.
Replace t with $\mathrm{ln}x$,
$\int \frac{\mathrm{ln}x}{x}dx=\frac{{\left(\mathrm{ln}x\right)}^{2}}{2}+c$.
The final solution of the integral is,
$\int \frac{x+\mathrm{ln}x}{x}dx=x+\frac{{\left(\mathrm{ln}x\right)}^{2}}{2}+C$.

Foreckije

$\int \frac{\mathrm{ln}\left(x\right)+x}{x}dx$
$=\int \left(\frac{\mathrm{ln}\left(x\right)}{x}+1\right)dx$
$=\int \frac{\mathrm{ln}\left(x\right)}{x}dx+\int 1dx$
$\int \frac{\mathrm{ln}\left(x\right)}{x}dx$
$=\int udu$
$=\frac{{u}^{2}}{2}$
$=\frac{{\mathrm{ln}}^{2}\left(x\right)}{2}$
$\int 1dx$
$\int \frac{\mathrm{ln}\left(x\right)}{x}dx+\int 1dx$
$=\frac{{\mathrm{ln}}^{2}\left(x\right)}{2}+x$
$\int \frac{\mathrm{ln}\left(x\right)+x}{x}dx$
$=\frac{{\mathrm{ln}}^{2}\left(x\right)}{2}+x+C$

karton

$\int \frac{x+\mathrm{ln}\left(x\right)}{x}dx$
Separate the fraction
$\int \frac{x}{x}+\frac{\mathrm{ln}\left(x\right)}{x}dx$
Divide
$\int 1+\frac{\mathrm{ln}\left(x\right)}{x}dx$
Use properties of integrals
$\int 1dx+\int \frac{\mathrm{ln}\left(x\right)}{x}dx$
Evaluate the integrals
$x+\frac{\mathrm{ln}\left(x{\right)}^{2}}{2}$
$x+\frac{\mathrm{ln}\left(x{\right)}^{2}}{2}+C$

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