What is the integral of (9+3x)/(1+x^2) ?

The integral of (9+3x)/(1+x^2) is 9arctan(x)+3/2 ln|1+x^2|+C

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integral of (9+3x)/(1+x^2)

\int \frac{9 + 3 x}{1 + x ^{2}} d x

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

Solution steps

\int \frac{9 + 3 x}{1 + x ^{2}} d x

Expand

\frac{9 + 3 x}{1 + x ^{2}} : \frac{9}{1 + x ^{2}} + \frac{3 x}{1 + x ^{2}}

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{9}{1 + x ^{2}} d x + \int \frac{3 x}{1 + x ^{2}} d x

\int \frac{9}{1 + x ^{2}} d x = 9 arctan (x)

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

= 9 arctan (x) + \frac{3}{2} ln 1 + x^{2}

Add a constant to the solution

= 9 arctan (x) + \frac{3}{2} ln 1 + x^{2} + C

\int_{0}^{\frac{π}{2}} - 4 cos^{5} (x) d x

\frac{d}{d x} (csc (3 x + 2))

\int cot (35 x) d x

x \to \infty lim (4 x - x^{3})

f (x) = \frac{4}{1 - 2 x ^{2}}

What is the integral of (9+3x)/(1+x^2) ?
The integral of (9+3x)/(1+x^2) is 9arctan(x)+3/2 ln|1+x^2|+C