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

The integral of (x^2ln(x))/((x^3+1)^2) is 1/9 (-(3ln(x))/(x^3+1)+ln|x^3|-ln|x^3+1|)+C

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integral of (x^2ln(x))/((x^3+1)^2)

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

\frac{1}{9} (- \frac{3 ln ( x )}{x ^{3} + 1} + ln x^{3} - ln x^{3} + 1) + C

Solution steps

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

Apply u-substitution

: \int \frac{ln ( u )}{9 ( u + 1 ) ^{2}} d u

= \int \frac{ln ( u )}{9 ( u + 1 ) ^{2}} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int \frac{ln ( u )}{( u + 1 ) ^{2}} d u

Apply Integration By Parts

: - \frac{ln ( u )}{u + 1} - \int - \frac{1}{u ( u + 1 )} d u

= \frac{1}{9} (- \frac{ln ( u )}{u + 1} - \int - \frac{1}{u ( u + 1 )} d u)

\int - \frac{1}{u ( u + 1 )} d u = - ln ∣ u ∣ + ln ∣ u + 1 ∣

= \frac{1}{9} (- \frac{ln ( u )}{u + 1} - (- ln ∣ u ∣ + ln ∣ u + 1 ∣))

Substitute back

u = x^{3}

= \frac{1}{9} (- \frac{ln ( x ^{3} )}{x ^{3} + 1} - (- ln x^{3} + ln x^{3} + 1))

Simplify

\frac{1}{9} (- \frac{ln ( x ^{3} )}{x ^{3} + 1} - (- ln x^{3} + ln x^{3} + 1)) : \frac{1}{9} (- \frac{3 ln ( x )}{x ^{3} + 1} + ln x^{3} - ln x^{3} + 1)

= \frac{1}{9} (- \frac{3 ln ( x )}{x ^{3} + 1} + ln x^{3} - ln x^{3} + 1)

Add a constant to the solution

= \frac{1}{9} (- \frac{3 ln ( x )}{x ^{3} + 1} + ln x^{3} - ln x^{3} + 1) + C

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What is the integral of (x^2ln(x))/((x^3+1)^2) ?
The integral of (x^2ln(x))/((x^3+1)^2) is 1/9 (-(3ln(x))/(x^3+1)+ln|x^3|-ln|x^3+1|)+C