What is the integral of (3-x)/(x^2+1) ?

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

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integral of (3-x)/(x^2+1)

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

3 arctan (x) - \frac{1}{2} ln x^{2} + 1 + C

Solution steps

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

Expand

\frac{3 - x}{x ^{2} + 1} : \frac{3}{x ^{2} + 1} - \frac{x}{x ^{2} + 1}

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

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

\int \frac{x}{x ^{2} + 1} d x = \frac{1}{2} ln x^{2} + 1

= 3 arctan (x) - \frac{1}{2} ln x^{2} + 1

Add a constant to the solution

= 3 arctan (x) - \frac{1}{2} ln x^{2} + 1 + C

\frac{\partial}{\partial x} (\frac{5 x y ^{2}}{x ^{2} + y ^{2}})

\frac{d}{d x} (x^{2} 3^{x})

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

y = f (x) = cos (3 x^{2})

6 y^{^{''}} - y^{^{'}} - y = 0, y (0) = 10, y^{^{'}} (0) = 0

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