What is the integral from 1 to 3 of 2/(4+(x-1)^2) ?

The integral from 1 to 3 of 2/(4+(x-1)^2) is pi/4

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integral from 1 to 3 of 2/(4+(x-1)^2)

\int_{1}^{3} \frac{2}{4 + ( x - 1 ) ^{2}} d x

\frac{π}{4}

Decimal Notation

0.78539 \dots

Solution steps

\int_{1}^{3} \frac{2}{4 + ( x - 1 ) ^{2}} d x

Take the constant out:

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

= 2 \cdot \int_{1}^{3} \frac{1}{4 + ( x - 1 ) ^{2}} d x

Apply u-substitution

: \int_{0}^{2} \frac{1}{4 + u ^{2}} d u

= 2 \cdot \int_{0}^{2} \frac{1}{4 + u ^{2}} d u

Apply Integral Substitution

= 2 \cdot \int_{0}^{1} \frac{1}{2 ( v ^{2} + 1 )} d v

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \int_{0}^{1} \frac{1}{v ^{2} + 1} d v

Use the common integral:

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

= 2 \cdot \frac{1}{2} [arctan (v)]_{0}^{1}

Simplify

2 \cdot \frac{1}{2} [arctan (v)]_{0}^{1} : [arctan (v)]_{0}^{1}

= [arctan (v)]_{0}^{1}

Compute the boundaries:

\frac{π}{4}

= \frac{π}{4}

y^{^{'}} = \frac{2 y}{x}

x \to 0 lim (\frac{x ^{2}}{x})

\frac{d}{d x} (2 sec (cos (x)) + 2 tan (2 x))

x \to 5 lim (\frac{x ^{2} + k x - 20}{x - 5})

(1, 128), (4, 161)

What is the integral from 1 to 3 of 2/(4+(x-1)^2) ?
The integral from 1 to 3 of 2/(4+(x-1)^2) is pi/4