Can a function have more than one different antiderivative? I keep getting different answers on some exercises and it drives me crazy, i double-checked my solving method and it's correct, yet i'm getting a different result.
Can a function have more than one different antiderivative? I keep getting different answers on some exercises and it drives me crazy, i double-checked my solving method and it's correct, yet i'm getting a different result.
I'm not taking into account the constant...
OK...here it is:
$\displaystyle \displaystyle\int\frac{x}{x^4-2x^2-1}\,dx$
my solution is $\displaystyle -\displaystyle\frac{1}{2(x^2-1)}$
The book only gives me these solutions where only 1 the correct one:
A) $\displaystyle \ln{\frac{x^2+2}{x^2+1}}$
B) $\displaystyle \ln{\frac{x^2+1}{x^2+2}}$
C) $\displaystyle \ln{|\frac{x^2-(\sqrt{2}+1)}{x^2+(\sqrt{2}-1)}|}$
D) $\displaystyle \frac{1}{4\sqrt{2}}\ln{|\frac{x^2-(\sqrt{2}+1)}{x^2+(\sqrt{2}-1)}|}$
E) $\displaystyle \ln{|\frac{x^2-1}{x^2+1}|}$
F) $\displaystyle \ln{|\frac{x^2+1}{x^2-1}|}$
And in fact the inclusion of the arbitrary constant is the reason why two superficially different answers can nevertheless be correct. eg. $\displaystyle \displaystyle {\int \frac{\sin (x)}{\cos^3 (x)} \, dx = \frac{1}{2} \tan^2 (x) + C}$, and this answer is equivalent to $\displaystyle \displaystyle {\frac{1}{2} \sec^2 (x) + K}$.
Your answer is quite wrong. Unfortunately, you have shown no working at all, so the errors you are making cannot be pointed out.
Read this: integrate x/(x^4 - 2x^2 - 1) - Wolfram|Alpha (be sure to click on Show steps).
Antiderivatives are nessecarily the same up-to the addition of a constant. For, if we suppose that two functions $\displaystyle f, g$ both have the same derivative, then it follows that
$\displaystyle \displaystyle
(f - g)' = 0
$
and by the Mean Value Theorem, we know this means $\displaystyle f - g$ is constant.
Hello, Utherr!
Your algebra is off . . .
$\displaystyle \displaystyle\int\frac{x}{x^4-2x^2-1}\,dx$
$\displaystyle (A)\;\ln\left[\dfrac{x^2+2}{x^2+1}\right] + C \qquad\qquad\quad (B)\;\ln\left[\dfrac{x^2+1}{x^2+2}\right] + C$
$\displaystyle (C)\; \ln\left|\dfrac{x^2-(\sqrt{2}+1)}{x^2+(\sqrt{2}-1)}\right| + C \qquad
(D)\;\dfrac{1}{4\sqrt{2}}\ln\left|\dfrac{x^2-(\sqrt{2}+1)}{x^2+(\sqrt{2}-1)}\right| + C$
$\displaystyle (E)\;\ln\left|\dfrac{x^2-1}{x^2+1}\right| + C \qquad\qquad\quad\; (F)\;\ln\left|\dfrac{x^2+1}{x^2-1}\right|+C$
The denominator is: .$\displaystyle x^4 - 2x^2 - 1 \;=\;x^4 - 2x^2 + 1 - 2 \;=\;(x^2-1)^2 - 2$
The integral is: .$\displaystyle \displaystyle{\int\frac{x\,dx}{(x^2-1)^2 - 2} }$
Let: .$\displaystyle u \,=\,x^2-1\quad\Rightarrow\quad du \,=\,2x\,dx \quad\Rightarrow\quad x\,dx \,=\,\frac{1}{2}du$
. . Substitute: .$\displaystyle \displaystyle{\int\frac{\frac{1}{2}du}{u^2-2} \;=\;\tfrac{1}{2}\int\frac{du}{u^2- (\sqrt{2})^2} }$
Formula: .$\displaystyle \displaystyle{\int \frac{du}{u^2-a^2} \;=\;\frac{1}{2a}\ln\left|\frac{u-a}{u+a}\right| + C}$
. . We have: .$\displaystyle \displaystyle{\frac{1}{2}\cdot\frac{1}{2\sqrt{2}}\ ln\left|\frac{u-\sqrt{2}}{u+\sqrt{2}}\right| + C \;=\;\frac{1}{4\sqrt{2}}\ln\left|\frac{u-\sqrt{2}}{u+\sqrt{2}}\right| + C }$
Back-substitute: .$\displaystyle \displaystyle{\frac{1}{4\sqrt{2}}\ln\left|\frac{(x ^2-1) - \sqrt{2}}{(x^2-1) + \sqrt{2}}\right| + C }$
. . . . . . . . . . .$\displaystyle \displaystyle{=\;\frac{1}{4\sqrt{2}}\ln\left|\frac {x-(\sqrt{2}+1)}{x + (\sqrt{2}-1)}\right| + C } \quad\hdots\;\text{answer (D)} $
Hello again Utherr!
A function can have different-looking antiderivatives
. . but they are all equivalent.
Here is a classic example: .$\displaystyle \displaystyle{\int \sin x\cos x\,dx}$
$\displaystyle \displaystyle{[1]\;\int\sin x(\cos x\,dx) }$
Let $\displaystyle u = \sin x \quad\Rightarrow\quad du = \cos x\,dx$
Substitute: .$\displaystyle \displaystyle{\int u\,du \:=\:\tfrac{1}{2}u^2 + C}$
Back-substitute: .$\displaystyle \boxed{\tfrac{1}{2}\sin^2\!x + C}$
$\displaystyle \displaystyle{[2]\;\int \cos x(\sin x\,dx) }$
Let $\displaystyle u = \cos x \quad\Rightarrow\quad du = -\sin x\,dx \quad\Rightarrow\quad \sin x\,dx = -du$
Substitute: .$\displaystyle \displaystyle{\int u (-du) \;=\;-\int u\,du \;=\;-\tfrac{1}{2}u^2+C }$
Back-substitute: .$\displaystyle \boxed{-\tfrac{1}{2}\cos^2\!x + C}$
$\displaystyle \displaystyle{[3]\;\int\sin x\cos x\,dx \;=\;\tfrac{1}{2}\int 2\sin x\cos x\,dx$
. . . $\displaystyle \displaystyle{=\;\tfrac{1}{2}\int\sin2x\,dx \;=\;\boxed{-\tfrac{1}{4}\cos^2\!2x + C} }$
You stated originally that you wanted to integrate $\displaystyle \int \frac{xdx}{x^4- 2x^2- 1}$ but here you are integrating $\displaystyle \int \frac{xdx}{x^4- 2x^2+ 1}$.
To integrate $\displaystyle \int\frac{xdx}{x^4- 2x^2- 1}$, write it as $\displaystyle \int \frac{xdx}{x^2- 2x^2+ 1- 2}= \int\frac{xdx}{(x^2-1)^2- 2}$. Now let $\displaystyle y= x^2- 1$ so that dy= 2xdx and the integral becomes $\displaystyle \frac{1}{2}\int \frac{dy}{y^2- 2}= \frac{1}{2}\int \frac{dy}{(y- \sqrt{2})(y+ \sqrt{2})}$ which can be integrated by "partial fractions".