What is the integral of (ln^2(y))/(3y+8yln^3(y)) ?

The integral of (ln^2(y))/(3y+8yln^3(y)) is 1/24 ln|3+8ln^3(y)|+C

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integral of (ln^2(y))/(3y+8yln^3(y))

\int \frac{ln ^{2} ( y )}{3 y + 8 y ln ^{3} ( y )} d y

\frac{1}{24} ln 3 + 8 ln^{3} (y) + C

Solution steps

\int \frac{ln ^{2} ( y )}{3 y + 8 y ln ^{3} ( y )} d y

Apply u-substitution

: \int \frac{u ^{2}}{3 + 8 u ^{3}} d u

= \int \frac{u ^{2}}{3 + 8 u ^{3}} d u

Apply u-substitution

: \int \frac{1}{24 v} d v

= \int \frac{1}{24 v} d v

Take the constant out:

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

= \frac{1}{24} \cdot \int \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{24} ln ∣ v ∣

Substitute back

= \frac{1}{24} ln 3 + 8 ln^{3} (y)

Add a constant to the solution

= \frac{1}{24} ln 3 + 8 ln^{3} (y) + C

\frac{( s + 5 )}{( s + 2 ) ( s + 3 )}

\int_{0}^{\frac{1}{3}} 2 x - 6 x^{2} d x

\int y^{7} 1 - y^{4} d y

\frac{d}{d x} (- x e^{2 x})

x \cdot y^{2} \cdot \frac{d y}{d x} = y^{3} - 4 \cdot x^{3}, y (1) = 2

What is the integral of (ln^2(y))/(3y+8yln^3(y)) ?
The integral of (ln^2(y))/(3y+8yln^3(y)) is 1/24 ln|3+8ln^3(y)|+C