What is the integral of (12x^3-1)/(3x^4-x+10) ?

The integral of (12x^3-1)/(3x^4-x+10) is ln|3x^4-x+10|+C

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integral of (12x^3-1)/(3x^4-x+10)

\int \frac{12 x ^{3} - 1}{3 x ^{4} - x + 10} d x

ln 3 x^{4} - x + 10 + C

Solution steps

\int \frac{12 x ^{3} - 1}{3 x ^{4} - x + 10} 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 = 3 x^{4} - x + 10

= ln 3 x^{4} - x + 10

Add a constant to the solution

= ln 3 x^{4} - x + 10 + C

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

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

3 cos (8 x), 3 - 3 cos (8 x), [0, \frac{π}{8}]

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

f (x) = \frac{1}{x - 4}

What is the integral of (12x^3-1)/(3x^4-x+10) ?
The integral of (12x^3-1)/(3x^4-x+10) is ln|3x^4-x+10|+C