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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Take the partial fraction of

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

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

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

\int \frac{1}{u} d u = ln ∣ u ∣

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

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

Substitute back

u = x^{2} + 1

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{1}{( x + 1 ) x ^{2} + 2 x - 8} d x

\frac{x ^{3} - 1}{x ^{2}}

\int \frac{3 x - 1}{x ^{2} + 9} d x

\frac{d}{d x} (ln (g (x)))

\int \frac{x}{( x ^{2} + 9 ) ^{2}} d x

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