How do you find the integral ${(ln x)}^{2}$ $(ln x)^2$ ?

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How do you find the integral ${(ln x)}^{2}$ $(ln x)^2$ ?

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Answer 1 · 2015-06-17T15:40:56

I found:
$\int {[ln (x)]}^{2} d x = x {ln}^{2} (x) - 2 x ln (x) + 2 x + c$ $int[ln(x)]^2dx=xln^2(x)-2xln(x)+2x+c$

Explanation:

I would try using Substitution and By Parts (twice):
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Answer 2 · 2015-06-17T21:11:22

I get the same answer as Gio,
$\int {(ln x)}^{2} d x = x {(ln x)}^{2} - 2 x ln x + 2 x + C$ $int(lnx)^2dx=x(lnx)^2-2xlnx+2x+C$

But the details of my solution are different.

Explanation:

$\int {(ln x)}^{2} d x$ $int(lnx)^2dx$

Use integration by parts:

Let $u = {(ln x)}^{2}$ $u = (lnx)^2$ and $d v = d x$ $dv = dx$ , so we have

$d u = \frac{2}{x} ln x d x$ $du = 2/x lnx dx$ and $v = x$ $v = x$ .

$\int {(ln x)}^{2} d x = x {(ln x)}^{2} - \int x \cdot \frac{2}{x} ln x d x$ $int(lnx)^2dx = x (lnx)^2 - int x* 2/xlnx dx$

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

$= x {(ln x)}^{2} - 2 [x ln x - x] + C$ $= x (lnx)^2 -2 [ xlnx - x]+C$

$= x {(ln x)}^{2} - 2 x ln x + 2 x + C$ $= x(lnx)^2-2xlnx+2x+C$

Note
If you don't know $\int ln x d x$ $int lnx dx$ , use integration by parts with

$u = ln x$ $u = lnx$ and $d v = d x$ $dv = dx$ .

General note
In general to integrate: $\int x^{r} ln x d x$ $int x^r ln x dx$ use $u = x^{r}$ $u=x^r$ and $d v = ln x$ $dv = lnx$ -- unless $r = - 1$ $r = -1$ in which case substitution $u = ln x$ $u = lnx$ is easier.

Answer 3 · 2015-06-18T17:43:59

$= \int ln x ln x d x$ $= intlnxlnxdx$

As funny as it sounds, I'm just going to do it like this. It's how I did it the first time I've seen this one. Let:
$u = ln x$ $u = lnx$
$d u = \frac{1}{x} d x$ $du = 1/xdx$
$d v = ln x d x$ $dv = lnxdx$
$v = x ln x - x$ $v = xlnx - x$

To do $\int ln x$ $int lnx$ :
Let:
$u = ln x$ $u = lnx$
$d v = 1 d x$ $dv = 1dx$
$d u = \frac{1}{x} d x$ $du = 1/xdx$
$v = x$ $v = x$

$x ln x - \int \frac{x}{x} d x = x ln x - x + C$ $xlnx - intx/xdx = xlnx - x + C$

Anyways, continuing on:

$\Rightarrow ln x (x ln x - x) - \int \frac{x ln x - x}{x} d x$ $=> lnx(xlnx - x) - int(xlnx-x)/xdx$

$= x {ln}^{2} x - x ln x - \int ln x - 1 d x$ $= xln^2x -xlnx - intlnx-1dx$

$= x {ln}^{2} x - x ln x - ((x ln x - x) - x)$ $= xln^2x -xlnx - ((xlnx-x)-x)$

$= x {ln}^{2} x - x ln x - x ln x + x + x + C$ $= xln^2x -xlnx - xlnx+x+x + C$

$= x {ln}^{2} x - 2 x ln x + 2 x + C$ $= xln^2x -2xlnx + 2x + C$

Cool, still got the same answer.

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How do you find the integral ${(ln x)}^{2}$ $(ln x)^2$ ?

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