What is the integral from 2 to 7 of 1/(25+(x-2)^2) ?

The integral from 2 to 7 of 1/(25+(x-2)^2) is pi/(20)

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integral from 2 to 7 of 1/(25+(x-2)^2)

\int_{2}^{7} \frac{1}{25 + ( x - 2 ) ^{2}} d x

\frac{π}{20}

Decimal Notation

0.15707 \dots

Solution steps

\int_{2}^{7} \frac{1}{25 + ( x - 2 ) ^{2}} d x

Apply u-substitution

: \int_{0}^{5} \frac{1}{25 + u ^{2}} d u

= \int_{0}^{5} \frac{1}{25 + u ^{2}} d u

Apply Integral Substitution

= \int_{0}^{1} \frac{1}{5 ( 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}{5} \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}{5} [arctan (v)]_{0}^{1}

Compute the boundaries:

\frac{π}{4}

= \frac{1}{5} \cdot \frac{π}{4}

Simplify

= \frac{π}{20}

\int (x^{\frac{3}{2}} - 5 x)^{2} d x

\frac{d}{d x} (sin^{3} (2 x - 3))

\frac{d y}{d t} + e^{t} y = 13 e^{t}

\frac{1}{4 + x ^{2}}, 0

\frac{d}{d x} (x^{3} - 4 x + 6)

What is the integral from 2 to 7 of 1/(25+(x-2)^2) ?
The integral from 2 to 7 of 1/(25+(x-2)^2) is pi/(20)