Ensino Superior ⇒ Primitiva Tópico resolvido

[subjectname]

Calcule a primitiva da função seguinte:

[tex3]\int \frac{x\cdot \ln x}{(x^2 +1)^2}\,dx[/tex3]

[jrneliodias]

Olá, Subjectname.

Note que

[tex3]\frac{x\cdot \ln x}{(x^2 +1)^2}\,\,=\,\,-\frac{1}{2}\cdot \frac{-2x}{(x^2 +1)^2}\cdot \ln x\,\,=\,\,-\frac{1}{2}\cdot \left[\frac{1}{(x^2 +1)}\right]'\cdot \ln x[/tex3]

Então se

[tex3]f(x)=-\frac{1}{2(x^2 +1)}[/tex3]

[tex3]g(x)=\ln x[/tex3]

[tex3]h(x)=f(x)\cdot g(x)[/tex3]

Se quisermos derivar [tex3]h[/tex3], teremos

[tex3]h'=f'\cdot g+f\cdot g'[/tex3]

Agora, se integrarmos:

[tex3]\int h'\,dx=\int f'\cdot g\,dx+\int f\cdot g'\, dx[/tex3]

Ou seja,

[tex3]\int h'\,dx=\int f'\cdot g\,dx+\int f\cdot g'\, dx[/tex3]

Assim,

[tex3]\int \frac{x\cdot \ln x}{(x^2 +1)^2}\,dx\,\,=\,\,-\frac{\ln x}{2(x^2 +1)}-\int \left(-\frac{1}{2\,x\,(x^2 +1)}\right) dx[/tex3]

[tex3]\int \frac{x\cdot \ln x}{(x^2 +1)^2}\,dx\,\,=\,\,-\frac{\ln x}{2(x^2 +1)}+\frac{1}{2}\,\int \frac{1}{x\,(x^2 +1)}\,dx[/tex3]

Porém,

[tex3]\frac{1}{x\,(x^2 +1)}\,\,=\,\,\frac{1}{x}-\frac{x}{x^2+1}\,\,=\,\,\frac{1}{x}-\frac{(x^2+1)'}{2(x^2+1)}[/tex3]

Substituindo,

[tex3]\int \frac{x\cdot \ln x}{(x^2 +1)^2}\,dx\,\,=\,\,-\frac{\ln x}{2(x^2 +1)}\,\,+\,\,\frac{1}{2}\,\int \frac{1}{x}\,dx\,\,-\,\,\frac{1}{4}\,\int \frac{(x^2+1)'}{x^2+1}\,dx[/tex3]

[tex3]\int \frac{x\cdot \ln x}{(x^2 +1)^2}\,dx\,\,=\,\,-\frac{\ln x}{2(x^2 +1)}\,\,+\,\,\frac{\ln x}{2}\,\,-\,\,\frac{\ln(x^2+1)}{4}+C[/tex3]

[tex3]\int \frac{x\cdot \ln x}{(x^2 +1)^2}\,dx\,\,=\,\,\frac{x^2\,\ln x}{2(x^2 +1)}\,\,-\,\,\frac{\ln(x^2+1)}{4}+C[/tex3]

Espero ter ajudado, abraço.