What is the integral of 1/(25+16x^2) ?

The integral of 1/(25+16x^2) is 1/20 arctan(4/5 x)+C

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integral of 1/(25+16x^2)

\int \frac{1}{25 + 16 x ^{2}} d x

\frac{1}{20} arctan (\frac{4}{5} x) + C

Solution steps

\int \frac{1}{25 + 16 x ^{2}} d x

Apply Integral Substitution

= \int \frac{1}{20 ( 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}{20} \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}{20} arctan (u)

Substitute back

u = \frac{4}{5} x

= \frac{1}{20} arctan (\frac{4}{5} x)

Add a constant to the solution

= \frac{1}{20} arctan (\frac{4}{5} x) + C

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

\frac{\partial}{\partial x} (x^{\frac{3}{4}} y^{\frac{1}{4}})

x \to 0 - lim (\frac{1}{e ^{\frac{1}{x}} - 1})

(x^{3} - x) (x^{2} + 2) (x^{2} + x - 1)

y^{^{''}} + 4 y^{^{'}} + 13 y = 0, y (0) = 2, y^{^{'}} (0) = - 2

What is the integral of 1/(25+16x^2) ?
The integral of 1/(25+16x^2) is 1/20 arctan(4/5 x)+C