Question

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

Answer 1

`\displaystyle \int (\ln(x))^2\, dx=x(\ln(x)^2-2\ln(x)+2)+C`

Explanation:

We want to evaluate the integral:

`\displaystyle \int\left(\ln x\right)^2 \, dx`

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

`\displaystyle =\int 1\cdot (\ln(x))^2\, dx`

Recall that IBP is given by:

`\displaystyle \int u\, dv = uv - \int v \, du`

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

`\displaystyle du = 2\ln x \cdot (1)/(x)\, dx`

Simplify:

`\displaystyle du=(2\ln(x))/(x)\, dx`

And:

`dv=(1)dx\Rightarrow v = x`

Therefore:

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

Simplify:

`\displaystyle =x\ln(x)^2-2 \int\ln(x)\, dx`

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

`\displaystyle du = (1)/(x) \, dx`

And:

`dv=(1)dx\Rightarrow v = x`

Thus:

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

Simplify:

`\displaystyle =x\ln(x)^2-2\left(x\ln(x)-\int (1)\, dx\right)`

Evaluate:

`=x\ln(x)^2-2(x\ln(x)-x)+C`

Simplify:

`\displaystyle =x\ln(x)^2-2x\ln(x)+2x`

Factor:

`=x(\ln(x)^2-2\ln(x)+2)`

Therefore:

`\displaystyle \int (\ln(x))^2\, dx=x(\ln(x)^2-2\ln(x)+2)+C`