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

The integral from-1 to 1 of 5/(1+25x^2) is 2arctan(5)

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

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

2 arctan (5)

Decimal Notation

2.74680 \dots

Solution steps

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

Take the constant out:

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

= 5 \cdot \int_{- 1}^{1} \frac{1}{1 + 25 x ^{2}} d x

Apply Integral Substitution

= 5 \cdot \int_{- 5}^{5} \frac{1}{5 ( 1 + u ^{2} )} d u

Take the constant out:

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

= 5 \cdot \frac{1}{5} \cdot \int_{- 5}^{5} \frac{1}{1 + u ^{2}} d u

Use the common integral:

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

= 5 \cdot \frac{1}{5} [arctan (u)]_{- 5}^{5}

Simplify

5 \cdot \frac{1}{5} [arctan (u)]_{- 5}^{5} : [arctan (u)]_{- 5}^{5}

= [arctan (u)]_{- 5}^{5}

Compute the boundaries:

2 arctan (5)

= 2 arctan (5)

\int_{y}^{y^{2}} (x + 2 y) d x

\int_{0}^{3} (1 + 2 x)^{2} d x

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

\int_{1}^{5} - x^{2} + 4 x d x

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

What is the integral from-1 to 1 of 5/(1+25x^2) ?
The integral from-1 to 1 of 5/(1+25x^2) is 2arctan(5)