What is the integral from 0 to 4 of 1/(49x^2+25) ?

The integral from 0 to 4 of 1/(49x^2+25) is 1/35 arctan(28/5)

Popular >

integral from 0 to 4 of 1/(49x^2+25)

\int_{0}^{4} \frac{1}{49 x ^{2} + 25} d x

\frac{1}{35} arctan (\frac{28}{5})

Decimal

0.03983 \dots

Solution steps

\int_{0}^{4} \frac{1}{49 x ^{2} + 25} d x

Apply Integral Substitution

= \int_{0}^{\frac{28}{5}} \frac{1}{35 ( 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}{35} \cdot \int_{0}^{\frac{28}{5}} \frac{1}{u ^{2} + 1} d u

Use the common integral:

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

= \frac{1}{35} [arctan (u)]_{0}^{\frac{28}{5}}

Compute the boundaries:

arctan (\frac{28}{5})

= \frac{1}{35} arctan (\frac{28}{5})

\int_{1}^{10} \frac{3}{x} d x

\int_{1}^{2} \frac{z ^{2} + 5}{z} d z

\int_{0}^{\frac{π}{2}} 2 x sin (2 x) d x

\int_{- 2}^{2} x + 2 d x

\int_{0}^{4} x e^{(- \frac{1}{2}) x^{2}} d x

What is the integral from 0 to 4 of 1/(49x^2+25) ?
The integral from 0 to 4 of 1/(49x^2+25) is 1/35 arctan(28/5)