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

Answer:

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

Step-by-step 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:

$\displaystyle du = 2\ln x \cdot \frac{1}{x}\, dx$

Simplify:

$\displaystyle du=\frac{2\ln(x)}{x}\, dx$

And:

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

Therefore:

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

Simplify:

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

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

$\displaystyle du = \frac{1}{x} \, dx$

And:

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

Thus:

$\displaystyle =x\ln(x)^2-2\left(x\ln(x)-\int (x)\frac{1}{x}\, dx\right)$

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:

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