What is the integral from 2 to 3 of 1/((x+1)ln(x+1)) ?

The integral from 2 to 3 of 1/((x+1)ln(x+1)) is ln(2ln(2))-ln(ln(3))

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integral from 2 to 3 of 1/((x+1)ln(x+1))

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

ln (2 ln (2)) - ln (ln (3))

Decimal Notation

0.23258 \dots

Solution steps

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

Apply u-substitution

: \int_{3}^{4} \frac{1}{u ln ( u )} d u

= \int_{3}^{4} \frac{1}{u ln ( u )} d u

Apply u-substitution

: \int_{l n (3)}^{2 l n (2)} \frac{1}{v} d v

= \int_{l n (3)}^{2 l n (2)} \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= [ln ∣ v ∣]_{l n (3)}^{2 l n (2)}

Compute the boundaries:

ln (2 ln (2)) - ln (ln (3))

= ln (2 ln (2)) - ln (ln (3))

\int \frac{csc ^{2} ( x )}{cot ^{4} ( x )} d x

f (x) = 10 x^{3} - 20 x + \frac{1}{5}

y^{^{''}} + y = \frac{1}{4} sin (3 t) - \frac{3}{4} sin (t)

\frac{d}{d x} (arccos (\frac{a x}{2}))

\int \frac{1}{x} - \frac{4}{x ^{2} + 1} d x

What is the integral from 2 to 3 of 1/((x+1)ln(x+1)) ?
The integral from 2 to 3 of 1/((x+1)ln(x+1)) is ln(2ln(2))-ln(ln(3))