What is the integral of (3x^2+4)/(0.3x^3+1.2x) ?

The integral of (3x^2+4)/(0.3x^3+1.2x) is 3.33333…ln|x|+3.33333…ln|0.3x^2+1.2|+C

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integral of (3x^2+4)/(0.3x^3+1.2x)

\int \frac{3 x ^{2} + 4}{0.3 x ^{3} + 1.2 x} d x

3.33333 \dots ln ∣ x ∣ + 3.33333 \dots ln 0.3 x^{2} + 1.2 + C

Solution steps

\int \frac{3 x ^{2} + 4}{0.3 x ^{3} + 1.2 x} d x

Factor

\frac{3 x ^{2} + 4}{0.3 x ^{3} + 1.2 x}

= \int \frac{3 x ^{2} + 4}{x ( 0.3 x ^{2} + 1.2 )} d x

Take the partial fraction of

\frac{3 x ^{2} + 4}{x ( 0.3 x ^{2} + 1.2 )} : \frac{10}{3 x} + \frac{2 x}{0.3 x ^{2} + 1.2}

= \int \frac{10}{3 x} + \frac{2 x}{0.3 x ^{2} + 1.2} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{10}{3 x} d x + \int \frac{2 x}{0.3 x ^{2} + 1.2} d x

\int \frac{10}{3 x} d x = \frac{10}{3} ln ∣ x ∣

\int \frac{2 x}{0.3 x ^{2} + 1.2} d x = 3.33333 \dots ln 0.3 x^{2} + 1.2

= \frac{10}{3} ln ∣ x ∣ + 3.33333 \dots ln 0.3 x^{2} + 1.2

Simplify

= 3.33333 \dots ln ∣ x ∣ + 3.33333 \dots ln 0.3 x^{2} + 1.2

Add a constant to the solution

= 3.33333 \dots ln ∣ x ∣ + 3.33333 \dots ln 0.3 x^{2} + 1.2 + C

f (x) = x^{6} tan (x)

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

t^{2} y^{^{''}} + 10 t y^{^{'}} + 8 y = 0

- 2 sec^{2} (x) tan (x)

\frac{d}{d x} (ln (cot (x)))

What is the integral of (3x^2+4)/(0.3x^3+1.2x) ?
The integral of (3x^2+4)/(0.3x^3+1.2x) is 3.33333…ln|x|+3.33333…ln|0.3x^2+1.2|+C