How do you integrate $\int 4 x {ln x}^{2} d x$ $int 4 x ln x^2 dx$ using integration by parts?

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Answer 1 · 2016-01-11T13:38:53

$\int u d v = 2 x^{2} ({ln x}^{2} - 1)$ $intu dv=2x^2(lnx^2-1)$

The Integration by Parts formula is shown by;

$\int u d v = u v - \int v d u$ $intu dv=u v-intv du$

To choose $u$ $u$ and $d v$ $dv$ , there are several things to be considered;

When answering this type of question, ensure that $u$ $u$ choosen can usually differentiates to zero, while $d v$ $dv$ is easy to integrate.

Also, choose $u$ $u$ in this order;
LIPET : L ogs, I nverse trig., P olynomial, E xponential, T rig.

In this question;
Let $u = {ln x}^{2}$ $u=lnx^2$ and $d v = 4 x d x$ $dv=4x dx$
Then, $d u = \frac{2}{x} d x$ $du=2/(x)dx$ and $v = 2 x^{2}$ $v=2x^2$

Using Integration by Parts formula;

$\int u d v = u v - \int v d u$ $intu dv=u v-intv du$

$\int u d v = {ln x}^{2} (2 x^{2}) - \int (2 x^{2}) (\frac{2}{x}) d x$ $intu dv=lnx^2(2x^2)-int(2x^2)(2/x)dx$

$\int u d v = {ln x}^{2} (2 x^{2}) - \int 4 x d x$ $intu dv=lnx^2(2x^2)-int4xdx$

$\int u d v = {ln x}^{2} (2 x^{2}) - 2 x^{2}$ $intu dv=lnx^2(2x^2)-2x^2$

$\int u d v = 2 x^{2} ({ln x}^{2} (1) - 1)$ $intu dv=2x^2(lnx^2(1)-1)$

$\int u d v = 2 x^{2} ({ln x}^{2} - 1)$ $intu dv=2x^2(lnx^2-1)$

How do you integrate ∫4xlnx2dxint 4 x ln x^2 dx using integration by parts?