What is the integral from 3 to 4 of 1/(x^2-6x+10) ?

The integral from 3 to 4 of 1/(x^2-6x+10) is pi/4

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integral from 3 to 4 of 1/(x^2-6x+10)

\int_{3}^{4} \frac{1}{x ^{2} - 6 x + 10} d x

\frac{π}{4}

Decimal Notation

0.78539 \dots

Solution steps

\int_{3}^{4} \frac{1}{x ^{2} - 6 x + 10} d x

Complete the square

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

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

Apply u-substitution

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

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

Use the common integral:

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

= [arctan (u)]_{0}^{1}

Compute the boundaries:

\frac{π}{4}

= \frac{π}{4}

\int_{0}^{3} \frac{t}{t ^{2} + 16} d t

\int_{3}^{4} 2 x d x

\int_{- 1}^{1} (x^{3}) d x

\int_{- 1}^{2} (x^{3} + x) d x

\int_{1}^{3} \frac{1}{x + 1} d x

What is the integral from 3 to 4 of 1/(x^2-6x+10) ?
The integral from 3 to 4 of 1/(x^2-6x+10) is pi/4