What is the integral from 3 to 5 of 2/(4+(x-3)^2) ?

The integral from 3 to 5 of 2/(4+(x-3)^2) is pi/4

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integral from 3 to 5 of 2/(4+(x-3)^2)

\int_{3}^{5} \frac{2}{4 + ( x - 3 ) ^{2}} d x

\frac{π}{4}

Decimal Notation

0.78539 \dots

Solution steps

\int_{3}^{5} \frac{2}{4 + ( x - 3 ) ^{2}} d x

Take the constant out:

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

= 2 \cdot \int_{3}^{5} \frac{1}{4 + ( x - 3 ) ^{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}

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

\frac{d}{d s} (\frac{1}{s + 2})

\int \frac{2}{1 - sin ( x )} d x

\int - 0.01 x d x

\int \frac{( x - 2 ) 1 - x ^{2}}{1 - x ^{2}} d x

What is the integral from 3 to 5 of 2/(4+(x-3)^2) ?
The integral from 3 to 5 of 2/(4+(x-3)^2) is pi/4