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

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

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

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

2 arctan (x) - 4 ln ∣ x ∣ + C

Solution steps

\int \frac{2}{1 + x ^{2}} - \frac{4}{x} 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{2}{1 + x ^{2}} d x - \int \frac{4}{x} d x

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

\int \frac{4}{x} d x = 4 ln ∣ x ∣

= 2 arctan (x) - 4 ln ∣ x ∣

Add a constant to the solution

= 2 arctan (x) - 4 ln ∣ x ∣ + C

f (x) = 3 x^{5} + 2 x^{2}

\int (4 x^{3} - 3 x^{2}) (4 x^{4} - 4 x^{3})^{5} d x

\int_{- 2}^{0} x^{3} - x^{2} - 6 x d x

\frac{\partial}{\partial x} (t^{- 3} \frac{x ^{2}}{2 t})

\int x^{- \frac{5}{3}} d x

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