What is the integral of (x+4)/(x^2+16) ?

The integral of (x+4)/(x^2+16) is 1/2 ln|x^2+16|+arctan(x/4)+C

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integral of (x+4)/(x^2+16)

\int \frac{x + 4}{x ^{2} + 16} d x

\frac{1}{2} ln x^{2} + 16 + arctan (\frac{x}{4}) + C

Solution steps

\int \frac{x + 4}{x ^{2} + 16} d x

Expand

\frac{x + 4}{x ^{2} + 16} : \frac{x}{x ^{2} + 16} + \frac{4}{x ^{2} + 16}

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} + 16} d x + \int \frac{4}{x ^{2} + 16} d x

\int \frac{x}{x ^{2} + 16} d x = \frac{1}{2} ln x^{2} + 16

\int \frac{4}{x ^{2} + 16} d x = arctan (\frac{x}{4})

= \frac{1}{2} ln x^{2} + 16 + arctan (\frac{x}{4})

Add a constant to the solution

= \frac{1}{2} ln x^{2} + 16 + arctan (\frac{x}{4}) + C

y^{^{''}} - 5 y^{^{'}} + 6 y = e^{t}

\int_{0}^{1} 9 t^{4} + 4 t^{2} d t

\int_{1}^{4} (x^{2} + x) d x

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

\int e^{x} - 1 d x

What is the integral of (x+4)/(x^2+16) ?
The integral of (x+4)/(x^2+16) is 1/2 ln|x^2+16|+arctan(x/4)+C