What is the integral of (x^4+3x^2+1)/(x^5+5x^3+5x) ?

The integral of (x^4+3x^2+1)/(x^5+5x^3+5x) is 1/5 ln|x^5+5x^3+5x|+C

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integral of (x^4+3x^2+1)/(x^5+5x^3+5x)

\int \frac{x ^{4} + 3 x ^{2} + 1}{x ^{5} + 5 x ^{3} + 5 x} d x

\frac{1}{5} ln x^{5} + 5 x^{3} + 5 x + C

Solution steps

\int \frac{x ^{4} + 3 x ^{2} + 1}{x ^{5} + 5 x ^{3} + 5 x} d x

Apply u-substitution

: \int \frac{1}{5 u} d u

= \int \frac{1}{5 u} d u

Take the constant out:

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

= \frac{1}{5} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{5} ln ∣ u ∣

Substitute back

u = x^{5} + 5 x^{3} + 5 x

= \frac{1}{5} ln x^{5} + 5 x^{3} + 5 x

Add a constant to the solution

= \frac{1}{5} ln x^{5} + 5 x^{3} + 5 x + C

\int e^{\frac{- x}{2}} d x

cot (x^{5})

5^{3^{x^{2}}}

x^{2} y^{^{'}} + 2 x y = 5 cos^{2} (x)

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

What is the integral of (x^4+3x^2+1)/(x^5+5x^3+5x) ?
The integral of (x^4+3x^2+1)/(x^5+5x^3+5x) is 1/5 ln|x^5+5x^3+5x|+C