What is the integral from 4 to 10 of 1/(36+(x-4)^2) ?

The integral from 4 to 10 of 1/(36+(x-4)^2) is pi/(24)

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integral from 4 to 10 of 1/(36+(x-4)^2)

\int_{4}^{10} \frac{1}{36 + ( x - 4 ) ^{2}} d x

\frac{π}{24}

Decimal Notation

0.13089 \dots

Solution steps

\int_{4}^{10} \frac{1}{36 + ( x - 4 ) ^{2}} d x

Apply u-substitution

: \int_{0}^{6} \frac{1}{36 + u ^{2}} d u

= \int_{0}^{6} \frac{1}{36 + u ^{2}} d u

Apply Integral Substitution

= \int_{0}^{1} \frac{1}{6 ( 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}{6} \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}{6} [arctan (v)]_{0}^{1}

Compute the boundaries:

\frac{π}{4}

= \frac{1}{6} \cdot \frac{π}{4}

Simplify

= \frac{π}{24}

\frac{d}{d x} (9 + x)

\frac{1}{9 s ^{2} + 1}

t \to 0 lim (e^{3 t})

y = 5 - 10 x + x^{3}, at x = - 1

What is the integral from 4 to 10 of 1/(36+(x-4)^2) ?
The integral from 4 to 10 of 1/(36+(x-4)^2) is pi/(24)