What is the integral from 2 to 5 of 4/(9+(x-2)^2) ?

The integral from 2 to 5 of 4/(9+(x-2)^2) is pi/3

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integral from 2 to 5 of 4/(9+(x-2)^2)

\int_{2}^{5} \frac{4}{9 + ( x - 2 ) ^{2}} d x

\frac{π}{3}

Decimal Notation

1.04719 \dots

Solution steps

\int_{2}^{5} \frac{4}{9 + ( x - 2 ) ^{2}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \int_{2}^{5} \frac{1}{9 + ( x - 2 ) ^{2}} d x

Apply u-substitution

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

= 4 \cdot \int_{0}^{3} \frac{1}{9 + u ^{2}} d u

Apply Integral Substitution

= 4 \cdot \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

= 4 \cdot \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)

= 4 \cdot \frac{1}{3} [arctan (v)]_{0}^{1}

Simplify

4 \cdot \frac{1}{3} [arctan (v)]_{0}^{1} : \frac{4}{3} [arctan (v)]_{0}^{1}

= \frac{4}{3} [arctan (v)]_{0}^{1}

Compute the boundaries:

\frac{π}{4}

= \frac{4}{3} \cdot \frac{π}{4}

Simplify

= \frac{π}{3}

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What is the integral from 2 to 5 of 4/(9+(x-2)^2) ?
The integral from 2 to 5 of 4/(9+(x-2)^2) is pi/3