What is the integral of [ 1/(x^2)+sec^2(pix)] ?

The integral of [ 1/(x^2)+sec^2(pix)] is -1/x+1/pi tan(pix)+C

Popular >

integral of [ 1/(x^2)+sec^2(pix)]

\int [\frac{1}{x ^{2}} + sec^{2} (π x)] d x

- \frac{1}{x} + \frac{1}{π} tan (π x) + C

Solution steps

\int \frac{1}{x ^{2}} + sec^{2} (π x) 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

= \int \frac{1}{x ^{2}} d x + \int sec^{2} (π x) d x

\int \frac{1}{x ^{2}} d x = - \frac{1}{x}

\int sec^{2} (π x) d x = \frac{1}{π} tan (π x)

= - \frac{1}{x} + \frac{1}{π} tan (π x)

Add a constant to the solution

= - \frac{1}{x} + \frac{1}{π} tan (π x) + C

\frac{\partial}{\partial x} ((x y)^{\frac{1}{2}})

n = 8 \sum \infty \frac{1}{4 ^{n}}

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

\frac{d}{d x} (2 x (x + 4)^{3})

\frac{\partial}{\partial x} (x^{2} + 3 y^{2} - 6 y + 4 x y)

What is the integral of [ 1/(x^2)+sec^2(pix)] ?
The integral of [ 1/(x^2)+sec^2(pix)] is -1/x+1/pi tan(pix)+C