What is the integral of (x^2+1)/(x^3+3x-16) ?

The integral of (x^2+1)/(x^3+3x-16) is 1/3 ln|x^3+3x-16|+C

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integral of (x^2+1)/(x^3+3x-16)

\int \frac{x ^{2} + 1}{x ^{3} + 3 x - 16} d x

\frac{1}{3} ln x^{3} + 3 x - 16 + C

Solution steps

\int \frac{x ^{2} + 1}{x ^{3} + 3 x - 16} d x

Apply u-substitution

: \int \frac{1}{3 u} d u

= \int \frac{1}{3 u} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{3} ln ∣ u ∣

Substitute back

u = x^{3} + 3 x - 16

= \frac{1}{3} ln x^{3} + 3 x - 16

Add a constant to the solution

= \frac{1}{3} ln x^{3} + 3 x - 16 + C

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\frac{1}{s ^{2} ( s + 1.1 )}

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

\int x x^{2} - 5 d x

What is the integral of (x^2+1)/(x^3+3x-16) ?
The integral of (x^2+1)/(x^3+3x-16) is 1/3 ln|x^3+3x-16|+C