What is the integral of 1000(1+(4x)/(2+x^2)) ?

The integral of 1000(1+(4x)/(2+x^2)) is 1000(x+2ln|2+x^2|)+C

integral of 1000(1+(4x)/(2+x^2))

\int 1000 (1 + \frac{4 x}{2 + x ^{2}}) d x

1000 (x + 2 ln 2 + x^{2}) + C

Solution steps

\int 1000 (1 + \frac{4 x}{2 + x ^{2}}) d x

Take the constant out:

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

= 1000 \cdot \int 1 + \frac{4 x}{2 + x ^{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

= 1000 (\int 1 d x + \int \frac{4 x}{2 + x ^{2}} d x)

\int 1 d x = x

\int \frac{4 x}{2 + x ^{2}} d x = 2 ln 2 + x^{2}

= 1000 (x + 2 ln 2 + x^{2})

Add a constant to the solution

= 1000 (x + 2 ln 2 + x^{2}) + C

\int \frac{r ln ( r ^{2} + 1 )}{r ^{2} + 1} d r

\int \frac{4 x}{( x + 2 ) ^{2}} d x

\int y^{2} 1 - y^{2} d y

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

\int (x + 1) (x^{2} + 2 x)^{3} d x

What is the integral of 1000(1+(4x)/(2+x^2)) ?
The integral of 1000(1+(4x)/(2+x^2)) is 1000(x+2ln|2+x^2|)+C