How do you integrate $\int x^{3} {ln x}^{2} d x$ $int x^3 ln x^2 dx$ using integration by parts?

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How do you integrate $\int x^{3} {ln x}^{2} d x$ $int x^3 ln x^2 dx$ using integration by parts?

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Answer 1 · 2016-10-05T17:47:01

$\frac{x^{4}}{8} (4 ln x - 1) + C$ $x^4/8(4lnx-1)+C$

Explanation:

Let $I = \int x^{3} {ln x}^{2} d x$ $I=intx^3lnx^2dx$

Using the Subst. $x^{2} = t, we have, 2 x d x = d t, or, x d x = \frac{1}{2} d t .$ $x^2=t", we have, "2xdx=dt, or, xdx=1/2dt.$ Hence,

$I = \int (x^{2} {ln x}^{2}) x d x = \frac{1}{2} \int t ln t d t .$ $I=int(x^2lnx^2)xdx=1/2inttlntdt.$

Now, to this end, we will use the Rule of I ntegration byParts arts :

$I B P : \int u v d t = u \int v d t - \int {\frac{d u}{d t} \int v d t} d t .$ $IBP : intuvdt=uintvdt-int{(du)/dtintvdt}dt.$

We take $u = ln t \Rightarrow \frac{d u}{d t} = \frac{1}{t}, &, v = t \Rightarrow \int v d t = \frac{t^{2}}{2}$ $u=lnt rArr (du)/dt=1/t, &, v=trArr intvdt=t^2/2$

$:. I=1/2[t^2/2lnt-int(1/t*t^2/2)dt]$

$=t^2/4lnt-1/4inttdt=t^2/4lnt-t^2/8$

Returning $t" by "x^2,$ we have,

$I=x^4/4ln(x^2)-1/8x^4$

Here, we use the Rule of Logarithm $: lnx^2=2lnx$ & get,

$I=x^4/2lnx-x^4/8=x^4/8(4lnx-1)$

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How do you integrate $\int x^{3} {ln x}^{2} d x$ $int x^3 ln x^2 dx$ using integration by parts?

1 Answer

Explanation:

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How do you integrate int x^3 ln x^2 dx ∫x3lnx2dxint x^3 ln x^2 dx using integration by parts?

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Explanation:

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How do you integrate $\int x^{3} {ln x}^{2} d x$ $int x^3 ln x^2 dx$ using integration by parts?