What is the integral from-1 to 1 of ln(ln(x+3)) ?

The integral from-1 to 1 of ln(ln(x+3)) is ln(16ln^2(2))-\Ei(ln(4))+\Ei(ln(2))

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integral from-1 to 1 of ln(ln(x+3))

\int_{- 1}^{1} ln (ln (x + 3)) d x

ln (16 ln^{2} (2)) - Ei (ln (4)) + Ei (ln (2))

Solution steps

\int_{- 1}^{1} ln (ln (x + 3)) d x

Apply u-substitution

: \int_{2}^{4} ln (ln (u)) d u

= \int_{2}^{4} ln (ln (u)) d u

Apply Integration By Parts

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

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

\int \frac{1}{ln ( u )} d u = Ei (ln (u))

= [u ln (ln (u)) - Ei (ln (u))]_{2}^{4}

Compute the boundaries:

4 ln (2 ln (2)) - Ei (ln (4)) - 2 ln (ln (2)) + Ei (ln (2))

= 4 ln (2 ln (2)) - Ei (ln (4)) - 2 ln (ln (2)) + Ei (ln (2))

Simplify

= ln (16 ln^{2} (2)) - Ei (ln (4)) + Ei (ln (2))

\frac{d x}{d y} = - \frac{( 4 y ^{2} + 6 x y )}{3 y ^{2} + 2 x}

\int_{0}^{2} (6 x^{2} - 4 x + 5) d x

y = e^{\frac{1}{2} x}

x \to a lim (\frac{f ( x )}{p ( x )})

y = 3 - x, y = 2 x^{2}, 0, 1

What is the integral from-1 to 1 of ln(ln(x+3)) ?
The integral from-1 to 1 of ln(ln(x+3)) is ln(16ln^2(2))-\Ei(ln(4))+\Ei(ln(2))