What is the integral from 1 to 4 of pi(ln(x))^2 ?

The integral from 1 to 4 of pi(ln(x))^2 is pi(16ln^2(2)-16ln(2)+6)

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integral from 1 to 4 of pi(ln(x))^2

\int_{1}^{4} π (ln (x))^{2} d x

π (16 ln^{2} (2) - 16 ln (2) + 6)

Decimal

8.15838 \dots

Solution steps

\int_{1}^{4} π (ln (x))^{2} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= π \cdot \int_{1}^{4} (ln (x))^{2} d x

Apply Integration By Parts

: [x ln^{2} (x) - \int 2 ln (x) d x]_{1}^{4}

= π [x ln^{2} (x) - \int 2 ln (x) d x]_{1}^{4}

\int 2 ln (x) d x = 2 (x ln (x) - x)

= π [x ln^{2} (x) - 2 (x ln (x) - x)]_{1}^{4}

Compute the boundaries:

16 ln^{2} (2) - 16 ln (2) + 6

= π (16 ln^{2} (2) - 16 ln (2) + 6)

\int_{n}^{n + 1} \frac{cos ( x )}{x} d x

\int_{0}^{1} - 2 y + 1 d y

\int_{5}^{6} x d x

\int_{0}^{\frac{π}{4}} t^{2} sin (2 t) d t

\int_{- 2}^{2} 2 x - 4 d x

What is the integral from 1 to 4 of pi(ln(x))^2 ?
The integral from 1 to 4 of pi(ln(x))^2 is pi(16ln^2(2)-16ln(2)+6)