Calculus 2 Master needed, evaluate the indefinite integral of: [tex]\int\( (lnx)^2} \, dx[/tex] So far I had applied the integration by parts and got: [tex](ln x )^2*x - \int\ x*((lnx)^2/2)}[/tex] do we do integration by parts again? also, do we simplify the x at my current stage? steps shown would be appreciated

Respuesta :

Answer:

[tex]\displaystyle \int (\ln(x))^2\, dx=x(\ln(x)^2-2\ln(x)+2)+C[/tex]

Step-by-step explanation:

We want to evaluate the integral:

[tex]\displaystyle \int\left(\ln x\right)^2 \, dx[/tex]

We can use Integration by Parts. Rewriting the integral yields:

[tex]\displaystyle =\int 1\cdot (\ln(x))^2\, dx[/tex]

Recall that IBP is given by:

[tex]\displaystyle \int u\, dv = uv - \int v \, du[/tex]

Let u be (ln(x))². And let dv be (1) dx. Therefore:

[tex]\displaystyle du = 2\ln x \cdot \frac{1}{x}\, dx[/tex]

Simplify:

[tex]\displaystyle du=\frac{2\ln(x)}{x}\, dx[/tex]

And:

[tex]dv=(1)dx\Rightarrow v = x[/tex]

Therefore:

[tex]\displaystyle \int (\ln(x))^2\, dx=\underbrace{\ln(x)^2}_{u}\underbrace{x}_{v}-\int\underbrace{x}_{v}\cdot \underbrace{\frac{2\ln(x)}{x}\, dx}_{du}[/tex]

Simplify:

[tex]\displaystyle =x\ln(x)^2-2 \int\ln(x)\, dx[/tex]

We can perform IBP again. Let u = ln(x) and v = 1. Hence:

[tex]\displaystyle du = \frac{1}{x} \, dx[/tex]

And:

[tex]dv=(1)dx\Rightarrow v = x[/tex]

Thus:

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

Simplify:

[tex]\displaystyle =x\ln(x)^2-2\left(x\ln(x)-\int (1)\, dx\right)[/tex]

Evaluate:

[tex]=x\ln(x)^2-2(x\ln(x)-x)+C[/tex]

Simplify:

[tex]\displaystyle =x\ln(x)^2-2x\ln(x)+2x[/tex]

Factor:

[tex]=x(\ln(x)^2-2\ln(x)+2)[/tex]

Therefore:

[tex]\displaystyle \int (\ln(x))^2\, dx=x(\ln(x)^2-2\ln(x)+2)+C[/tex]

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