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

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

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

integral

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

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

Solution steps

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

Apply Integral Substitution

= \int \frac{u + 3}{u ^{2} + 1} d u

Expand

\frac{u + 3}{u ^{2} + 1} : \frac{u}{u ^{2} + 1} + \frac{3}{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

= \int \frac{u}{u ^{2} + 1} d u + \int \frac{3}{u ^{2} + 1} d u

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

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

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

Substitute back

u = \frac{x}{3}

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

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

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

Add a constant to the solution

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

in t e g r a l \frac{1}{cos ^{2} ( y )}

in t e g r a l \frac{x ^{2}}{x}

in t e g r a l \frac{2 sin ( x )}{( 5 + 3 cos ( x ) )}

in t e g r a l v^{2}

in t e g r a l \frac{( x ^{2} + x + 1 )}{( x ^{2} - x )}

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