What is the integral of (4x^2+3)/(0.4x^3+0.9x) ?

The integral of (4x^2+3)/(0.4x^3+0.9x) is 3.33333…ln|x|+3.33333…ln|0.4x^2+0.9|+C

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integral of (4x^2+3)/(0.4x^3+0.9x)

\int \frac{4 x ^{2} + 3}{0.4 x ^{3} + 0.9 x} d x

3.33333 \dots ln ∣ x ∣ + 3.33333 \dots ln 0.4 x^{2} + 0.9 + C

Solution steps

\int \frac{4 x ^{2} + 3}{0.4 x ^{3} + 0.9 x} d x

Factor

\frac{4 x ^{2} + 3}{0.4 x ^{3} + 0.9 x}

= \int \frac{4 x ^{2} + 3}{x ( 0.4 x ^{2} + 0.9 )} d x

Take the partial fraction of

\frac{4 x ^{2} + 3}{x ( 0.4 x ^{2} + 0.9 )} : \frac{10}{3 x} + \frac{2.66666 \dots x}{0.4 x ^{2} + 0.9}

= \int \frac{10}{3 x} + \frac{2.66666 \dots x}{0.4 x ^{2} + 0.9} 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{2.66666 \dots x}{0.4 x ^{2} + 0.9} d x

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

\int \frac{2.66666 \dots x}{0.4 x ^{2} + 0.9} d x = 3.33333 \dots ln 0.4 x^{2} + 0.9

= \frac{10}{3} ln ∣ x ∣ + 3.33333 \dots ln 0.4 x^{2} + 0.9

Simplify

= 3.33333 \dots ln ∣ x ∣ + 3.33333 \dots ln 0.4 x^{2} + 0.9

Add a constant to the solution

= 3.33333 \dots ln ∣ x ∣ + 3.33333 \dots ln 0.4 x^{2} + 0.9 + C

\frac{d}{d t} (\frac{t ^{3} - 2 t ^{2} + 2 t - 4}{t ^{2} + 2})

sin (θ)

\int \frac{1}{x ^{2} - 10 x} d x

y^{^{'}} = y^{2} + 3

f (x) = 2 x^{2} + 3 x - 2, at a = - 1

What is the integral of (4x^2+3)/(0.4x^3+0.9x) ?
The integral of (4x^2+3)/(0.4x^3+0.9x) is 3.33333…ln|x|+3.33333…ln|0.4x^2+0.9|+C