# How do you integrate h(t)=(t+1)/(t^2+2t+2) using the quotient rule?

Nov 4, 2016

$\int \frac{t + 1}{{t}^{2} + 2 t + 2} \mathrm{dx} = \left(\frac{1}{2}\right) \ln \left({t}^{2} + 2 t + 2\right) + C$

#### Explanation:

The quotient rule does not pertain to integrals; it pertains to derivatives.

This integral is best integrated by u substitution:

let $u = {t}^{2} + 2 t + 2$, then $\mathrm{du} = \left(2 t + 2\right) \mathrm{dx}$, therefore, we shall substitute $\left(\frac{1}{2}\right) \mathrm{du}$ for $\left(t + 1\right) \mathrm{dx}$:

$\int \frac{t + 1}{{t}^{2} + 2 t + 2} \mathrm{dx} = \left(\frac{1}{2}\right) \int \frac{\mathrm{du}}{u}$

$\int \frac{t + 1}{{t}^{2} + 2 t + 2} \mathrm{dx} = \left(\frac{1}{2}\right) \ln | u | + C$

Reverse the substitution for u:

$\int \frac{t + 1}{{t}^{2} + 2 t + 2} \mathrm{dx} = \left(\frac{1}{2}\right) \ln | {t}^{2} + 2 t + 2 | + C$

Because ${t}^{2} + 2 t + 2$ can never be less than or equal to zero, we may drop the absolute value:

$\int \frac{t + 1}{{t}^{2} + 2 t + 2} \mathrm{dx} = \left(\frac{1}{2}\right) \ln \left({t}^{2} + 2 t + 2\right) + C$