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

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

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

\int \frac{4 x ^{3} + 3}{x ^{4} + 3 x} d x

ln x^{4} + 3 x + C

Solution steps

\int \frac{4 x ^{3} + 3}{x ^{4} + 3 x} d x

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= ln ∣ u ∣

Substitute back

u = x^{4} + 3 x

= ln x^{4} + 3 x

Add a constant to the solution

= ln x^{4} + 3 x + C

y^{^{''}} + 2 y^{^{'}} + y = 19 e^{- t}

\frac{d}{d x} (- 3 e^{x})

\frac{d ^{2}}{d x ^{2}} (x^{\frac{6}{7}} + 4 x)

\frac{\partial}{\partial x} (k!)

\frac{d}{d x} (\frac{6}{1 + e ^{- x}})

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