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

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

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

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

ln (2) + ln ∣ t + 1 ∣ - ln ∣ t + 3 ∣

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int_{1}^{t} \frac{1}{3 + x ^{2} + 4 x} d x

Take the partial fraction of

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

= 2 \cdot \int_{1}^{t} \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

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

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

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

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

Simplify

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

= ln (2) + ln ∣ t + 1 ∣ - ln ∣ t + 3 ∣

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