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

The integral of (x^3+2x)/(x^4+4x^2+3) is 1/4 ln|x^4+4x^2+3|+C

Popular >

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

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

\frac{1}{4} ln x^{4} + 4 x^{2} + 3 + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{4 u} d u

= \int \frac{1}{4 u} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{4} ln ∣ u ∣

Substitute back

u = x^{4} + 4 x^{2} + 3

= \frac{1}{4} ln x^{4} + 4 x^{2} + 3

Add a constant to the solution

= \frac{1}{4} ln x^{4} + 4 x^{2} + 3 + C

\frac{d y}{d t} + 0.7 t y = 7 t, y (0) = 7

y^{^{''}} + 2 y^{^{'}} + 2 y = sin (2 x)

\frac{d}{d x} (\frac{50}{x ^{9}})

\int \frac{8}{4 - x ^{2}} d x

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

What is the integral of (x^3+2x)/(x^4+4x^2+3) ?
The integral of (x^3+2x)/(x^4+4x^2+3) is 1/4 ln|x^4+4x^2+3|+C