What is the integral from 3 to 5 of 1/(4+(x-3)^2) ?

The integral from 3 to 5 of 1/(4+(x-3)^2) is pi/8

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integral from 3 to 5 of 1/(4+(x-3)^2)

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

\frac{π}{8}

Decimal Notation

0.39269 \dots

Solution steps

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

Apply u-substitution

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

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

Apply Integral Substitution

= \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

= \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)

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

Compute the boundaries:

\frac{π}{4}

= \frac{1}{2} \cdot \frac{π}{4}

Simplify

= \frac{π}{8}

y^{^{'}} + 3 y = e^{x} y^{2}

x \to 4 + lim (\frac{x + 5}{x - 4})

9 y^{^{''}} + y = 0

y^{^{''}} + 4 y = t^{2} + 3 e^{t}

\int_{0}^{π} cos^{6} (x) d x

What is the integral from 3 to 5 of 1/(4+(x-3)^2) ?
The integral from 3 to 5 of 1/(4+(x-3)^2) is pi/8