Multiple choice question of indefinite integral, $\int \frac{x + 9}{x^3 + 9x} dx$.

Question

If $\int \frac{x + 9}{x^3 + 9x} dx = k\arctan(mx) + n\ln (x) + p \ln (x^2 + 9) + c$, then $(m+n)/(k+p) = $
(A) 6
(B) -8
(C) -3
(D) 4

I tried solving it by differentiating the R.H.S. but couldn't arrive at the answer.

Answer 1

The derivative of the right hand side is $$ \frac{km}{1+m^2x^2}+\frac{n}{x}+\frac{2px}{x^2+9} $$ which should equal $$ \frac{x + 9}{x^3 + 9x} $$ Thus you need $m=1/3$ and then you can go on.

Answer 2

\begin{align} \int \frac{x + 9}{x^3 + 9x} dx &= \int \frac{x}{(x^2 + 9)x} dx + \int \frac{9}{(x^2 + 9)x} dx \\ \\ &= \int \frac{dx}{x^2 + 9} + \int \left(\frac{A}{x} + \frac{Bx+C}{x^2 + 9}\right) dx \\ \\ &= \int \frac{dx}{x^2 + 9} + \int \left(\frac{1}{x} + \frac{-x}{x^2 + 9}\right) dx \\ \\ &= \frac{1}{9}\int \frac{dx}{\left(x/3\right)^2 + 1} + \int \frac{dx}{x} - \int\frac{x}{x^2 + 9}dx \\ \\ &= \frac{1}{3}\arctan(x/3) + \ln x - \frac{1}{2}\ln(x^2+9) + c \end{align}

$$\int \frac{x + 9}{x^3 + 9x} dx = k\arctan(mx) + n\ln (x) + p \ln (x^2 + 9) + c$$ $$\frac{m+n}{k+p} = \frac{1/3+1}{1/3-1/2}=\frac{4/3}{-1/6}=-8$$

Answer 3

Try to separate the denominator like this

$$\frac{x+9}{x^3+9x}=\frac{x+9}{x(x^2+9)}$$ $$\qquad\qquad\ \,=\frac Ax+\frac{Bx+C}{x^2+9}$$

Then construct an equality for $A,B,C$ and we have

$$Ax^2+9A+Bx^2+Cx=x+9$$

Here we get $A=1,\ B=-1,\ C=1$, and that gives us

$$\int\frac{x+9}{x^3+9x}\,dx\ =\ \int\frac1x+\frac1{x^2+9}-\frac x{x^2+9}\,dx$$

Could you continue with this?

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Note that \begin{align*} &\int\frac{x}{x^2+9}\,dx\\ =\ \frac12&\int\frac{du}{u+9}\qquad(u=x^2,\ du=2x\,dx)\\ =\ \frac12&\ln(u+9)+c\\ =\ \frac12&\ln(x^2+9)+c \end{align*}