What is the integral from 5 to 9 of x/(x^2-10x+41) ?

The integral from 5 to 9 of x/(x^2-10x+41) is 1/2 ln(2)+(5pi)/(16)

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integral from 5 to 9 of x/(x^2-10x+41)

\int_{5}^{9} \frac{x}{x ^{2} - 10 x + 41} d x

\frac{1}{2} ln (2) + \frac{5 π}{16}

Decimal Notation

1.32832 \dots

Solution steps

\int_{5}^{9} \frac{x}{x ^{2} - 10 x + 41} d x

Complete the square

x^{2} - 10 x + 41 : (x - 5)^{2} + 16

= \int_{5}^{9} \frac{x}{( x - 5 ) ^{2} + 16} d x

Apply u-substitution

: \int_{0}^{4} \frac{u + 5}{u ^{2} + 16} d u

= \int_{0}^{4} \frac{u + 5}{u ^{2} + 16} d u

Expand

\frac{u + 5}{u ^{2} + 16} : \frac{u}{u ^{2} + 16} + \frac{5}{u ^{2} + 16}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{4} \frac{u}{u ^{2} + 16} d u + \int_{0}^{4} \frac{5}{u ^{2} + 16} d u

\int_{0}^{4} \frac{u}{u ^{2} + 16} d u = \frac{1}{2} ln (2)

\int_{0}^{4} \frac{5}{u ^{2} + 16} d u = \frac{5 π}{16}

= \frac{1}{2} ln (2) + \frac{5 π}{16}

f (x) = 2 t^{- \frac{1}{4}}

sin (5 ln (x))

f (x) = \frac{2 x}{x ^{2} + 1}, at x = 0

x \to \infty lim (- 7 x)

\int \frac{1}{x ^{2} + 14 x} d x

What is the integral from 5 to 9 of x/(x^2-10x+41) ?
The integral from 5 to 9 of x/(x^2-10x+41) is 1/2 ln(2)+(5pi)/(16)