1. ## partial fraction integration

(x+1) / [x(x²+1)]

i worked it out and got.....

ln|x|-(1/2)ln|x²+1| + c

is this right or wrong? my book has something to do with arctan in it!? i dont see how!

2. You can always check an integration by differentiating, and seeing if you get back to the original integrand. Do you in this case?

3. Here is a solution (it does not use partial fractions,
so you would still have to do it yourself, of course):

$\displaystyle \displaystyle I = \int \frac{x+1}{x(x^2+1)}\;{dx} = \int\frac{1+\frac{1}{x}}{x^2+1}}\;{dx}.$

Let $\displaystyle t= \frac{1}{x}$, then $\displaystyle dx = -x^2\;{dt} = -\frac{1}{t^2}\;{dt},$ thus:

\displaystyle \displaystyle \begin{aligned} I & = -\int \frac{1+t}{t^2\left(\frac{1}{t^2}+1\right)}\;{dt} = -\int\frac{1+t}{1+t^2}\;{dt} \\& = -\int \frac{1}{1+t^2}\;{dt}-\int\frac{t}{1+t^2}\;{dt} \\& = -\tan^{-1}{t}-\frac{1}{2}\int\frac{2t}{1+t^2}\;{dt} \\& = -\tan^{-1}{t}-\frac{1}{2}\ln\left(1+t^2\right)+k \\& = -\tan^{-1}{\frac{1}{x}}-\frac{1}{2}\ln\left(1+\frac{1}{x^2}\right)+k.\end{ aligned}

4. If you want to use Partial Fractions:

$\displaystyle \displaystyle \frac{A}{x} + \frac{Bx + C}{x^2 + 1} = \frac{x+1}{x(x^2 + 1)}$

$\displaystyle \displaystyle \frac{A(x^2+1) + (Bx+C)x}{x(x^2 + 1)} = \frac{x+1}{x(x^2 + 1)}$

$\displaystyle \displaystyle A(x^2+1) + (Bx+C)x = x + 1$

$\displaystyle \displaystyle Ax^2 + A + Bx^2 + Cx = x + 1$

$\displaystyle \displaystyle (A + B)x^2 + Cx + A = 0x^2 + x + 1$.

So $\displaystyle \displaystyle A = 1, B = -1, C = 1$.

So $\displaystyle \displaystyle \frac{x + 1}{x(x^2 + 1)} = \frac{1}{x} + \frac{-x + 1}{x^2 + 1} = \frac{1}{x} - \frac{x}{x^2 + 1} + \frac{1}{x^2 + 1}$.

Therefore $\displaystyle \displaystyle \int{\frac{x+1}{x(x^2 + 1)}\,dx} = \int{\frac{1}{x} - \frac{x}{x^2 + 1} + \frac{1}{x^2 + 1}\,dx}$.

The first term is easy to integrate, the second needs a $\displaystyle \displaystyle u$ substitution, the third needs a trigonometric substitution.