Problems from Integration by parts

Development:

We have to choose a function that is $u(x)$ and other one $v(x)$, so that $\ln (x)$ is expressed like $\ln (x)=u(x)\cdot v^{\prime }(x)$.

We choose in this case: $$u=\ln (x);dv=1\cdot dx$$

And we have $$du={\displaystyle \frac{1}{x}};v={\displaystyle \int 1\cdot dx=x}$$

We can now apply the integration by parts formula, and we have:

$$\int \ln (x)dx=\int \ln (x)\cdot 1dx=x\cdot \ln (x)-\int x\cdot {\displaystyle \frac{1}{x}}dx=$$ $$=x\cdot \ln (x)-\int 1dx=x\cdot \ln (x)-x+C$$

When we have to integrate logarithms it is often useful to take $u(x)=\ln (x)$ since its derivative can generally be simplified with other terms in the integral.

Solution:

${\displaystyle \int \ln (x)dx=x\cdot \ln (x)-x+C}$

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