What is the integral of log_{e}(x^2) ?

The integral of log_{e}(x^2) is 2xln(x)-2x+C

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integral of log_{e}(x^2)

\int lo g_{e} (x^{2}) d x

2 x ln (x) - 2 x + C

Solution steps

\int lo g_{e} (x^{2}) d x

Apply Integration By Parts

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

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

\int 2 d x = 2 x

= x ln (x^{2}) - 2 x

Apply log rule

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x),

assuming

x \geq 0

= 2 x ln (x) - 2 x

Add a constant to the solution

= 2 x ln (x) - 2 x + C

e^{5 t} (- 2 cos (t) + 12 sin (t))

\frac{\partial}{\partial x} (y 1 + x^{7})

\int \frac{11 x ^{2}}{( 144 + x ^{2} ) ^{2}} d x

\frac{d}{d x} (x^{6} ln (7 x))

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

What is the integral of log_{e}(x^2) ?
The integral of log_{e}(x^2) is 2xln(x)-2x+C