What is the integral from 0 to 1 of (5/(7+7x^2)) ?

The integral from 0 to 1 of (5/(7+7x^2)) is (5pi)/(28)

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integral from 0 to 1 of (5/(7+7x^2))

\int_{0}^{1} (\frac{5}{7 + 7 x ^{2}}) d x

\frac{5 π}{28}

Decimal

0.56099 \dots

Solution steps

\int_{0}^{1} \frac{5}{7 + 7 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_{0}^{1} \frac{1}{7 + 7 x ^{2}} d x

\frac{1}{7 + 7 x ^{2}} = \frac{1}{7} \frac{1}{1 + x ^{2}}

= 5 \cdot \int_{0}^{1} \frac{1}{7} \cdot \frac{1}{1 + x ^{2}} d x

Take the constant out:

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

= 5 \cdot \frac{1}{7} \cdot \int_{0}^{1} \frac{1}{1 + x ^{2}} d x

Use the common integral:

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

= 5 \cdot \frac{1}{7} [arctan (x)]_{0}^{1}

Simplify

5 \cdot \frac{1}{7} [arctan (x)]_{0}^{1} : \frac{5}{7} [arctan (x)]_{0}^{1}

= \frac{5}{7} [arctan (x)]_{0}^{1}

Compute the boundaries:

\frac{π}{4}

= \frac{5}{7} \cdot \frac{π}{4}

Simplify

= \frac{5 π}{28}

\int_{0}^{2} x x d x

\int_{1}^{2} \frac{1 + y}{1 + y ^{2}} d y

\int_{1}^{4} 7 x d x

\int_{0}^{1} 5 (1 - x)^{4} d x

\int_{- 1}^{9} x^{\frac{1}{5}} - x d x

What is the integral from 0 to 1 of (5/(7+7x^2)) ?
The integral from 0 to 1 of (5/(7+7x^2)) is (5pi)/(28)