What is the integral from 0 to 1 of 1/(x^2+4x+3) ?

The integral from 0 to 1 of 1/(x^2+4x+3) is (-ln(2)+ln(3))/2

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integral from 0 to 1 of 1/(x^2+4x+3)

\int_{0}^{1} \frac{1}{x ^{2} + 4 x + 3} d x

\frac{- ln ( 2 ) + ln ( 3 )}{2}

Decimal

0.20273 \dots

Solution steps

\int_{0}^{1} \frac{1}{x ^{2} + 4 x + 3} d x

Take the partial fraction of

\frac{1}{x ^{2} + 4 x + 3} : \frac{1}{2 ( x + 1 )} - \frac{1}{2 ( x + 3 )}

= \int_{0}^{1} \frac{1}{2 ( x + 1 )} - \frac{1}{2 ( x + 3 )} d x

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{1}{2 ( x + 1 )} d x - \int_{0}^{1} \frac{1}{2 ( x + 3 )} d x

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

\int_{0}^{1} \frac{1}{2 ( x + 3 )} d x = ln (2) - \frac{1}{2} ln (3)

= \frac{1}{2} ln (2) - (ln (2) - \frac{1}{2} ln (3))

Simplify

\frac{1}{2} ln (2) - (ln (2) - \frac{1}{2} ln (3)) : \frac{- ln ( 2 ) + ln ( 3 )}{2}

= \frac{- ln ( 2 ) + ln ( 3 )}{2}

\int_{0}^{1} 1 + (2 y)^{2} d y

\int_{- 5}^{3} x^{2} + x - 12 d x

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

\int_{0}^{5.4} 378 - 17 t^{2} d t

\int_{0}^{4} (x^{2} - 4 x + 4)^{2} d x

What is the integral from 0 to 1 of 1/(x^2+4x+3) ?
The integral from 0 to 1 of 1/(x^2+4x+3) is (-ln(2)+ln(3))/2