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

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Answer 1 · 2016-01-17T13:05:43

$\int {ln x}^{2} d x = x \cdot {ln x}^{2} - 2 x + C$ $int ln x^2 dx=x* ln x^2 - 2x + C$

The formula $\int u \cdot d v = u \cdot v - \int v \cdot d u$ $int u *dv= u*v-int v* du$

the given $\int {ln x}^{2} d x$ $int ln x^2 dx$
Let $u = {ln x}^{2}$ $u=ln x^2$
$d v = d x$ $dv =dx$

$v = x$ $v=x$

$d u = 2 x \cdot \frac{d x}{x^{2}} = 2 \cdot \frac{d x}{x}$ $du=2x*dx/x^2=2* dx/x$

$\int {ln x}^{2} d x = x \cdot {ln x}^{2} - \int x \cdot 2 \cdot \frac{d x}{x} + C$ $int ln x^2 dx=x*ln x^2 - int x*2*dx/x+C$

$\int {ln x}^{2} d x = x \cdot {ln x}^{2} - \int 2 \cdot d x + C$ $int ln x^2 dx=x*ln x^2 - int 2*dx+C$

$\int {ln x}^{2} d x = x \cdot {ln x}^{2} - 2 x + C$ $int ln x^2 dx=x* ln x^2 - 2x + C$

How do you integrate int ln x^2 dx ∫lnx2dxint ln x^2 dx using integration by parts?