Question

Calculus 2 Master needed, evaluate the indefinite integral of: %20%5C%2C%20dx" id="TexFormula1" title="\int\( (lnx)^2} \, dx" alt="\int\( (lnx)^2} \, dx" align="absmiddle" class="latex-formula"> 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:

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

Step-by-step explanation:

So we have the indefinite integral:

$\int (\ln(x))^2dx$

This is the same thing as:

$=\int 1\cdot (\ln(x))^2dx$

So, let's do integration by parts.

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

$u=(\ln(x))^2\\\text{Find du. Use the chain rule.}\\\frac{du}{dx}=2(\ln(x))\cdot\frac{1}{x}$

Simplify:

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

And:

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

Therefore:

$\int (\ln(x))^2dx=x\ln(x)^2-\int(x)(\frac{2\ln(x)}{x})dx$

The x cancel:

$=x\ln(x)^2-\int2\ln(x)dx$

Move the 2 to the front:

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

(I'm not exactly sure how you got what you got. Perhaps you differentiated incorrectly?)

Now, let's use integrations by parts again for the integral. Similarly, let's put a 1 in front:

$=x\ln(x)^2-2\int 1\cdot\ln(x)dx$

Let u be ln(x) and let dv be (1)dx. Thus:

$u=\ln(x)\\du=\frac{1}{x}dx$

And:

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

So:

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

Simplify the integral:

$=x\ln(x)^2-2(x\ln(x)-\int (1)dx)$

Evaluate:

$=x\ln(x)^2-2(x\ln(x)-x)$

Now, we just have to simplify :)

Distribute the -2:

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

And if preferred, we can factor out a x:

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

And, of course, don't forget about the constant of integration!

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

And we are done :)