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

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

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

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

\frac{1}{2} ln x^{2} + 1 + 4 arctan (x) + C

Solution steps

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

Expand

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

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

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

= \frac{1}{2} ln x^{2} + 1 + 4 arctan (x)

Add a constant to the solution

= \frac{1}{2} ln x^{2} + 1 + 4 arctan (x) + C

f (t) = t \cdot sin (3 t)

\int sinh (x) + C d x

\frac{\partial}{\partial y} (x^{n} sin (\frac{y}{x ^{2}}))

f (x) = e^{3 t s i n (2 t)}

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

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