What is the integral from 0 to 5/4 of 1/(25+16x^2) ?

The integral from 0 to 5/4 of 1/(25+16x^2) is pi/(80)

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integral from 0 to 5/4 of 1/(25+16x^2)

\int_{0}^{\frac{5}{4}} \frac{1}{25 + 16 x ^{2}} d x

\frac{π}{80}

Decimal Notation

0.03926 \dots

Solution steps

\int_{0}^{\frac{5}{4}} \frac{1}{25 + 16 x ^{2}} d x

Apply Integral Substitution

= \int_{0}^{1} \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_{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}{20} [arctan (u)]_{0}^{1}

Compute the boundaries:

\frac{π}{4}

= \frac{1}{20} \cdot \frac{π}{4}

Simplify

= \frac{π}{80}

\frac{d}{d x} (y ln (x))

y = \frac{4 ln ( 3 x )}{5 + 2 x}

6 cos (t), at t = - \frac{π}{4}

2 s + 1

f (x) = x^{4} - 4 x^{3} + 2 x^{2} - 6, at x = 1

What is the integral from 0 to 5/4 of 1/(25+16x^2) ?
The integral from 0 to 5/4 of 1/(25+16x^2) is pi/(80)