What is the integral of 1/(x^2+49) ?

The integral of 1/(x^2+49) is 1/7 arctan(x/7)+C

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integral of 1/(x^2+49)

\int \frac{1}{x ^{2} + 49} d x

\frac{1}{7} arctan (\frac{x}{7}) + C

Solution steps

\int \frac{1}{x ^{2} + 49} d x

Apply Integral Substitution

= \int \frac{1}{7 ( u ^{2} + 1 )} d u

Take the constant out:

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

= \frac{1}{7} \cdot \int \frac{1}{u ^{2} + 1} d u

Use the common integral:

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

= \frac{1}{7} arctan (u)

Substitute back

u = \frac{x}{7}

= \frac{1}{7} arctan (\frac{x}{7})

Add a constant to the solution

= \frac{1}{7} arctan (\frac{x}{7}) + C

x^{3}, 3 x^{2} - 4

x^{3} y^{^{'}} + 3 x^{2} y = \frac{1}{x}, y (1) = 3

\frac{\partial}{\partial x} (x sin (y) sin (z))

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

\int (- 3 x + 4) e - x d x

What is the integral of 1/(x^2+49) ?
The integral of 1/(x^2+49) is 1/7 arctan(x/7)+C