What is the integral from 5 to 8 of 1/(9+(x-5)^2) ?

The integral from 5 to 8 of 1/(9+(x-5)^2) is pi/(12)

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integral from 5 to 8 of 1/(9+(x-5)^2)

\int_{5}^{8} \frac{1}{9 + ( x - 5 ) ^{2}} d x

\frac{π}{12}

Decimal Notation

0.26179 \dots

Solution steps

\int_{5}^{8} \frac{1}{9 + ( x - 5 ) ^{2}} d x

Apply u-substitution

: \int_{0}^{3} \frac{1}{9 + u ^{2}} d u

= \int_{0}^{3} \frac{1}{9 + u ^{2}} d u

Apply Integral Substitution

= \int_{0}^{1} \frac{1}{3 ( 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}{3} \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}{3} [arctan (v)]_{0}^{1}

Compute the boundaries:

\frac{π}{4}

= \frac{1}{3} \cdot \frac{π}{4}

Simplify

= \frac{π}{12}

n = 1 \sum \infty 2^{n + 1} - 2^{n}

n = 0 \sum \infty (sin (8))^{n}

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

\int (\frac{x ^{2} - 4}{x}) d x

\frac{d y}{d x}, x = y^{4}

What is the integral from 5 to 8 of 1/(9+(x-5)^2) ?
The integral from 5 to 8 of 1/(9+(x-5)^2) is pi/(12)