What is the integral from 0 to 5/2 of 1/(25+4x^2) ?

The integral from 0 to 5/2 of 1/(25+4x^2) is pi/(40)

Popular >

integral from 0 to 5/2 of 1/(25+4x^2)

\int_{0}^{\frac{5}{2}} \frac{1}{25 + 4 x ^{2}} d x

\frac{π}{40}

Solution steps

\int_{0}^{\frac{5}{2}} \frac{1}{25 + 4 x ^{2}} d x

Apply Integral Substitution

= \int_{0}^{1} \frac{1}{10 ( 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}{10} \cdot \int_{0}^{1} \frac{1}{u ^{2} + 1} d u

Use the common integral:

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

= \frac{1}{10} [arctan (u)]_{0}^{1}

Compute the boundaries:

\frac{π}{4}

= \frac{1}{10} \cdot \frac{π}{4}

Simplify

= \frac{π}{40}

\int 3 x^{2} - 12 x + 12 d x

\int_{1}^{4} (2 x + x) d x

\int ln (n)

\int \frac{2 e ^{3 t}}{1 + e ^{6 t}} d t

x \to 2 + lim (\frac{∣ x ∣ + 2}{x - 2})

What is the integral from 0 to 5/2 of 1/(25+4x^2) ?
The integral from 0 to 5/2 of 1/(25+4x^2) is pi/(40)