What is the integral from 2 to 4 of 2/(x^2-6x+10) ?

The integral from 2 to 4 of 2/(x^2-6x+10) is pi

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integral from 2 to 4 of 2/(x^2-6x+10)

\int_{2}^{4} \frac{2}{x ^{2} - 6 x + 10} d x

π

Decimal Notation

3.14159 \dots

Solution steps

\int_{2}^{4} \frac{2}{x ^{2} - 6 x + 10} d x

Take the constant out:

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

= 2 \cdot \int_{2}^{4} \frac{1}{x ^{2} - 6 x + 10} d x

Complete the square

x^{2} - 6 x + 10 : (x - 3)^{2} + 1

= 2 \cdot \int_{2}^{4} \frac{1}{( x - 3 ) ^{2} + 1} d x

Apply u-substitution

: \int_{- 1}^{1} \frac{1}{u ^{2} + 1} d u

= 2 \cdot \int_{- 1}^{1} \frac{1}{u ^{2} + 1} d u

Use the common integral:

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

= 2 [arctan (u)]_{- 1}^{1}

Compute the boundaries:

\frac{π}{2}

= 2 \cdot \frac{π}{2}

Simplify

= π

(- 2.1) (4.2)

f (x) = (x^{2} + 7) (5 - x), at x = 0

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

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

\int x sin^{2} (9 x) d x

What is the integral from 2 to 4 of 2/(x^2-6x+10) ?
The integral from 2 to 4 of 2/(x^2-6x+10) is pi