What is the integral of 1/(x^4(x^3+1)^2) ?

The integral of 1/(x^4(x^3+1)^2) is 1/3 (2ln|x^3+1|-1/(x^3+1)-2ln|x^3|-1/(x^3))+C

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integral of 1/(x^4(x^3+1)^2)

\int \frac{1}{x ^{4} ( x ^{3} + 1 ) ^{2}} d x

\frac{1}{3} (2 ln x^{3} + 1 - \frac{1}{x ^{3} + 1} - 2 ln x^{3} - \frac{1}{x ^{3}}) + C

Solution steps

\int \frac{1}{x ^{4} ( x ^{3} + 1 ) ^{2}} d x

Apply u-substitution

: \int \frac{1}{3 u ^{2} ( u - 1 ) ^{2}} d u

= \int \frac{1}{3 u ^{2} ( 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}{3} \cdot \int \frac{1}{u ^{2} ( u - 1 ) ^{2}} d u

Take the partial fraction of

\frac{1}{u ^{2} ( u - 1 ) ^{2}} : \frac{2}{u} + \frac{1}{u ^{2}} - \frac{2}{u - 1} + \frac{1}{( u - 1 ) ^{2}}

= \frac{1}{3} \cdot \int \frac{2}{u} + \frac{1}{u ^{2}} - \frac{2}{u - 1} + \frac{1}{( u - 1 ) ^{2}} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{3} (\int \frac{2}{u} d u + \int \frac{1}{u ^{2}} d u - \int \frac{2}{u - 1} d u + \int \frac{1}{( u - 1 ) ^{2}} d u)

\int \frac{2}{u} d u = 2 ln ∣ u ∣

\int \frac{1}{u ^{2}} d u = - \frac{1}{u}

\int \frac{2}{u - 1} d u = 2 ln ∣ u - 1 ∣

\int \frac{1}{( u - 1 ) ^{2}} d u = - \frac{1}{u - 1}

= \frac{1}{3} (2 ln ∣ u ∣ - \frac{1}{u} - 2 ln ∣ u - 1 ∣ - \frac{1}{u - 1})

Substitute back

u = x^{3} + 1

= \frac{1}{3} (2 ln x^{3} + 1 - \frac{1}{x ^{3} + 1} - 2 ln x^{3} + 1 - 1 - \frac{1}{x ^{3} + 1 - 1})

Simplify

= \frac{1}{3} (2 ln x^{3} + 1 - \frac{1}{x ^{3} + 1} - 2 ln x^{3} - \frac{1}{x ^{3}})

Add a constant to the solution

= \frac{1}{3} (2 ln x^{3} + 1 - \frac{1}{x ^{3} + 1} - 2 ln x^{3} - \frac{1}{x ^{3}}) + C

\int x \cdot x^{2} + 3 d x

\int x^{6} (x^{7} - 3)^{\frac{3}{4}} d x

x \to 0 + lim ((ln (\frac{1}{x}))^{x})

\frac{d}{d x} (10 cos (\frac{3}{10} x))

n = 3 \sum \infty \frac{n + 7}{n !}

What is the integral of 1/(x^4(x^3+1)^2) ?
The integral of 1/(x^4(x^3+1)^2) is 1/3 (2ln|x^3+1|-1/(x^3+1)-2ln|x^3|-1/(x^3))+C