What is the integral of (x+2)/(x^2+25) ?

The integral of (x+2)/(x^2+25) is 1/2 ln|(x^2)/(25)+1|+2/5 arctan(x/5)+C

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integral of (x+2)/(x^2+25)

\int \frac{x + 2}{x ^{2} + 25} d x

\frac{1}{2} ln \frac{x ^{2}}{25} + 1 + \frac{2}{5} arctan (\frac{x}{5}) + C

Solution steps

\int \frac{x + 2}{x ^{2} + 25} d x

Apply Integral Substitution

= \int \frac{5 u + 2}{5 ( u ^{2} + 1 )} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{5} \cdot \int \frac{5 u + 2}{u ^{2} + 1} d u

Expand

\frac{5 u + 2}{u ^{2} + 1} : \frac{5 u}{u ^{2} + 1} + \frac{2}{u ^{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

= \frac{1}{5} (\int \frac{5 u}{u ^{2} + 1} d u + \int \frac{2}{u ^{2} + 1} d u)

\int \frac{5 u}{u ^{2} + 1} d u = \frac{5}{2} ln u^{2} + 1

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

= \frac{1}{5} (\frac{5}{2} ln u^{2} + 1 + 2 arctan (u))

Substitute back

u = \frac{x}{5}

= \frac{1}{5} (\frac{5}{2} ln (\frac{x}{5})^{2} + 1 + 2 arctan (\frac{x}{5}))

Simplify

\frac{1}{5} (\frac{5}{2} ln (\frac{x}{5})^{2} + 1 + 2 arctan (\frac{x}{5})) : \frac{1}{2} ln \frac{x ^{2}}{25} + 1 + \frac{2}{5} arctan (\frac{x}{5})

= \frac{1}{2} ln \frac{x ^{2}}{25} + 1 + \frac{2}{5} arctan (\frac{x}{5})

Add a constant to the solution

= \frac{1}{2} ln \frac{x ^{2}}{25} + 1 + \frac{2}{5} arctan (\frac{x}{5}) + C

f (x) = \frac{10 x}{1 + x ^{2}}, (3, 3)

x^{5} + y^{4} x^{6} + 5 y^{^{'}} = 0

(- 5) (7.8)

x \to - \infty lim (0. 5^{x})

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

What is the integral of (x+2)/(x^2+25) ?
The integral of (x+2)/(x^2+25) is 1/2 ln|(x^2)/(25)+1|+2/5 arctan(x/5)+C