What is the integral from-1 to 1 of (x+5)/(x^2+2x+5) ?

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

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integral from-1 to 1 of (x+5)/(x^2+2x+5)

\int_{- 1}^{1} \frac{x + 5}{x ^{2} + 2 x + 5} d x

\frac{ln ( 2 ) + π}{2}

Decimal Notation

1.91736 \dots

Solution steps

\int_{- 1}^{1} \frac{x + 5}{x ^{2} + 2 x + 5} d x

Complete the square

x^{2} + 2 x + 5 : (x + 1)^{2} + 4

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

Apply u-substitution

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

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

Apply Integral Substitution

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

Expand

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

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

\int_{0}^{1} \frac{v}{v ^{2} + 1} d v = \frac{1}{2} ln (2)

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

= \frac{1}{2} ln (2) + \frac{π}{2}

Simplify

\frac{1}{2} ln (2) + \frac{π}{2} : \frac{ln ( 2 ) + π}{2}

= \frac{ln ( 2 ) + π}{2}

\int_{0}^{4} \frac{1}{x ^{3}} d x

\int_{0}^{\frac{π}{5}} x cos (5 x) d x

\int_{0}^{7} tan (x) d x

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

\int_{0}^{\frac{π}{2}} sin (x - y) d y

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