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

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

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

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

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

Solution steps

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

Apply u-substitution

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

Take the constant out:

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

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

Take the partial fraction of

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

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

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

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

= \frac{1}{2} (- \frac{1}{u} - arctan (u))

Substitute back

u = x^{2}

= \frac{1}{2} (- \frac{1}{x ^{2}} - arctan (x^{2}))

Add a constant to the solution

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

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

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in t e g r a l x^{\frac{3}{2}}

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