What is the integral of xln(3x^2) ?

The integral of xln(3x^2) is 1/2 ln(3)x^2+x^2ln(x)-(x^2)/2+C

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integral of xln(3x^2)

\int x ln (3 x^{2}) d x

\frac{1}{2} ln (3) x^{2} + x^{2} ln (x) - \frac{x ^{2}}{2} + C

Solution steps

\int x ln (3 x^{2}) d x

Apply Integration By Parts

: \frac{1}{2} x^{2} ln (3 x^{2}) - \int x d x

= \frac{1}{2} x^{2} ln (3 x^{2}) - \int x d x

\int x d x = \frac{x ^{2}}{2}

= \frac{1}{2} x^{2} ln (3 x^{2}) - \frac{x ^{2}}{2}

Simplify

\frac{1}{2} x^{2} ln (3 x^{2}) - \frac{x ^{2}}{2} : \frac{1}{2} ln (3) x^{2} + x^{2} ln (x) - \frac{x ^{2}}{2}

= \frac{1}{2} ln (3) x^{2} + x^{2} ln (x) - \frac{x ^{2}}{2}

Add a constant to the solution

= \frac{1}{2} ln (3) x^{2} + x^{2} ln (x) - \frac{x ^{2}}{2} + C

\frac{1}{x} d x + d y = 0

arctan (e^{4 x})

arccos (4 x^{9})

y = \frac{x ^{3}}{1 - x ^{2}}

What is the integral of xln(3x^2) ?
The integral of xln(3x^2) is 1/2 ln(3)x^2+x^2ln(x)-(x^2)/2+C