What is the integral from 5 to 9 of 1/(16+(x-5)^2) ?

The integral from 5 to 9 of 1/(16+(x-5)^2) is pi/(16)

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integral from 5 to 9 of 1/(16+(x-5)^2)

\int_{5}^{9} \frac{1}{16 + ( x - 5 ) ^{2}} d x

\frac{π}{16}

Decimal Notation

0.19634 \dots

Solution steps

\int_{5}^{9} \frac{1}{16 + ( x - 5 ) ^{2}} d x

Apply u-substitution

: \int_{0}^{4} \frac{1}{16 + u ^{2}} d u

= \int_{0}^{4} \frac{1}{16 + u ^{2}} d u

Apply Integral Substitution

= \int_{0}^{1} \frac{1}{4 ( 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}{4} \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}{4} [arctan (v)]_{0}^{1}

Compute the boundaries:

\frac{π}{4}

= \frac{1}{4} \cdot \frac{π}{4}

Simplify

= \frac{π}{16}

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

\frac{d}{d x} ((8 x - 1)^{3} (2 x + 1)^{4})

\int x \cdot cos (a x) d x

\frac{4 x + 3}{x - 1} (2.11)

y = 3 x^{2} - 3 x + 4

What is the integral from 5 to 9 of 1/(16+(x-5)^2) ?
The integral from 5 to 9 of 1/(16+(x-5)^2) is pi/(16)