What is the integral of (x+1)/(x^2+16) ?

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

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integral of (x+1)/(x^2+16)

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

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

Solution steps

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

Expand

\frac{x + 1}{x ^{2} + 16} : \frac{x}{x ^{2} + 16} + \frac{1}{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{1}{x ^{2} + 16} d x

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

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

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

Add a constant to the solution

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

f (x) = x (x - 4)

y^{^{''}} + 2 y^{^{'}} + y = 10 e^{- t}

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

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

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

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