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

The integral of (1+x^2)/k is 1/k (x+(x^3)/3)+C

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integral of (1+x^2)/k

integral

\frac{1 + x ^{2}}{k}

\frac{1}{k} (x + \frac{x ^{3}}{3}) + C

Solution steps

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

Take the constant out:

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

= \frac{1}{k} \cdot \int 1 + x^{2} d x

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

\int 1 d x = x

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

= \frac{1}{k} (x + \frac{x ^{3}}{3})

Add a constant to the solution

= \frac{1}{k} (x + \frac{x ^{3}}{3}) + C

(1 - x) e^{x}

\frac{1}{( x ^{2} - 2 x )}

4 x + 5

x (a^{2} - x^{2})^{\frac{1}{2}}

\frac{1}{( x ( x ^{4} + 1 ) )}

What is the integral of (1+x^2)/k ?
The integral of (1+x^2)/k is 1/k (x+(x^3)/3)+C