What is the integral from 1 to e of ln(x^2) ?

The integral from 1 to e of ln(x^2) is 2

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integral from 1 to e of ln(x^2)

\int_{1}^{e} ln (x^{2}) d x

2

Solution steps

\int_{1}^{e} ln (x^{2}) d x

Apply Integration By Parts

: [x ln (x^{2}) - \int 2 d x]_{1}^{e}

= [x ln (x^{2}) - \int 2 d x]_{1}^{e}

\int 2 d x = 2 x

= [x ln (x^{2}) - 2 x]_{1}^{e}

Apply log rule

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

assuming

x \geq 0

= [2 x ln (x) - 2 x]_{1}^{e}

Compute the boundaries:

2

= 2

\int_{110}^{160} \frac{160}{3 x ^{2}} d x

\frac{d}{d x} (\frac{3}{x - 2})

y = cos (a^{3} + x^{3})

\frac{d y}{d x} = - 5 e^{x + y}

\int_{1}^{4} (3 x^{2} + x) d x

What is the integral from 1 to e of ln(x^2) ?
The integral from 1 to e of ln(x^2) is 2