What is the integral from 6 to 7 of x/(x^2+4x+20) ?

The integral from 6 to 7 of x/(x^2+4x+20) is 1/2 (ln(97/16)-ln(5)-arctan(9/4)+arctan(2))

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integral from 6 to 7 of x/(x^2+4x+20)

\int_{6}^{7} \frac{x}{x ^{2} + 4 x + 20} d x

\frac{1}{2} (ln (\frac{97}{16}) - ln (5) - arctan (\frac{9}{4}) + arctan (2))

Decimal Notation

0.07363 \dots

Solution steps

\int_{6}^{7} \frac{x}{x ^{2} + 4 x + 20} d x

Complete the square

x^{2} + 4 x + 20 : (x + 2)^{2} + 16

= \int_{6}^{7} \frac{x}{( x + 2 ) ^{2} + 16} d x

Apply u-substitution

: \int_{8}^{9} \frac{u - 2}{u ^{2} + 16} d u

= \int_{8}^{9} \frac{u - 2}{u ^{2} + 16} d u

Apply Integral Substitution

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

Expand

\frac{2 v - 1}{v ^{2} + 1} : \frac{2 v}{v ^{2} + 1} - \frac{1}{v ^{2} + 1}

Apply the Sum Rule:

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

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

\int_{2}^{\frac{9}{4}} \frac{2 v}{v ^{2} + 1} d v = ln (\frac{97}{16}) - ln (5)

\int_{2}^{\frac{9}{4}} \frac{1}{v ^{2} + 1} d v = arctan (\frac{9}{4}) - arctan (2)

= \frac{1}{2} (ln (\frac{97}{16}) - ln (5) - (arctan (\frac{9}{4}) - arctan (2)))

Simplify

= \frac{1}{2} (ln (\frac{97}{16}) - ln (5) - arctan (\frac{9}{4}) + arctan (2))

\int \frac{x ^{3}}{x ^{2} + 2} d x

\int \frac{cos ( a x )}{1 - sin ( a x )} d x

n = 1 \sum \infty \frac{x ^{n}}{4 n - 1}

y = e^{(4 x + 3)}

\frac{∣ x ∣}{2 - x ^{2}}

What is the integral from 6 to 7 of x/(x^2+4x+20) ?
The integral from 6 to 7 of x/(x^2+4x+20) is 1/2 (ln(97/16)-ln(5)-arctan(9/4)+arctan(2))