What is the integral of (4+3x)/(1+x^2) ?

The integral of (4+3x)/(1+x^2) is 4arctan(x)+3/2 ln|1+x^2|+C

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integral of (4+3x)/(1+x^2)

\int \frac{4 + 3 x}{1 + x ^{2}} d x

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

Solution steps

\int \frac{4 + 3 x}{1 + x ^{2}} d x

Expand

\frac{4 + 3 x}{1 + x ^{2}} : \frac{4}{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{4}{1 + x ^{2}} d x + \int \frac{3 x}{1 + x ^{2}} d x

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

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

= 4 arctan (x) + \frac{3}{2} ln 1 + x^{2}

Add a constant to the solution

= 4 arctan (x) + \frac{3}{2} ln 1 + x^{2} + C

\int \frac{e ^{- 2 x}}{x ^{2}} d x

\int_{0}^{\frac{π}{3}} sec^{4} (x) d x

f (x) = 8 - x^{2}

\int u^{1} d u

y^{(\frac{1}{2})} \frac{d y}{d x} + y^{(\frac{3}{2})} = 1, y (0) = 9

What is the integral of (4+3x)/(1+x^2) ?
The integral of (4+3x)/(1+x^2) is 4arctan(x)+3/2 ln|1+x^2|+C