What is the integral of (60x^2+20x)/(10x^3+5x^2+3) ?

The integral of (60x^2+20x)/(10x^3+5x^2+3) is 2ln|10x^3+5x^2+3|+C

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integral of (60x^2+20x)/(10x^3+5x^2+3)

\int \frac{60 x ^{2} + 20 x}{10 x ^{3} + 5 x ^{2} + 3} d x

2 ln 10 x^{3} + 5 x^{2} + 3 + C

Solution steps

\int \frac{60 x ^{2} + 20 x}{10 x ^{3} + 5 x ^{2} + 3} d x

Apply u-substitution

: \int \frac{2}{u} d u

= \int \frac{2}{u} d u

Take the constant out:

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

= 2 \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 ln ∣ u ∣

Substitute back

u = 10 x^{3} + 5 x^{2} + 3

= 2 ln 10 x^{3} + 5 x^{2} + 3

Add a constant to the solution

= 2 ln 10 x^{3} + 5 x^{2} + 3 + C

y = x x

\int \frac{1}{( x ^{2} + 64 ) ^{2}} d x

\frac{d}{d x} (e^{\frac{x}{10}})

x \to o - lim (4)

\frac{d}{d x} ((x^{2} + 1)^{6})

What is the integral of (60x^2+20x)/(10x^3+5x^2+3) ?
The integral of (60x^2+20x)/(10x^3+5x^2+3) is 2ln|10x^3+5x^2+3|+C