# Thread: Antiderivative(s) of a function

1. ## Antiderivative(s) of a function

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.

2. All integrable functions have an infinite number of antiderivatives. This is why you have to include the integration constant after performing any indefinite integration.

3. I'm not taking into account the constant...

OK...here it is:

$\displaystyle\int\frac{x}{x^4-2x^2-1}\,dx$

my solution is $-\displaystyle\frac{1}{2(x^2-1)}$

The book only gives me these solutions where only 1 the correct one:

A) $\ln{\frac{x^2+2}{x^2+1}}$
B) $\ln{\frac{x^2+1}{x^2+2}}$
C) $\ln{|\frac{x^2-(\sqrt{2}+1)}{x^2+(\sqrt{2}-1)}|}$
D) $\frac{1}{4\sqrt{2}}\ln{|\frac{x^2-(\sqrt{2}+1)}{x^2+(\sqrt{2}-1)}|}$
E) $\ln{|\frac{x^2-1}{x^2+1}|}$
F) $\ln{|\frac{x^2+1}{x^2-1}|}$

4. Originally Posted by Prove It
All integrable functions have an infinite number of antiderivatives. This is why you have to include the integration constant after performing any indefinite integration.
And in fact the inclusion of the arbitrary constant is the reason why two superficially different answers can nevertheless be correct. eg. $\displaystyle {\int \frac{\sin (x)}{\cos^3 (x)} \, dx = \frac{1}{2} \tan^2 (x) + C}$, and this answer is equivalent to $\displaystyle {\frac{1}{2} \sec^2 (x) + K}$.

5. Originally Posted by Utherr
I'm not taking into account the constant...

OK...here it is:

$\displaystyle\int\frac{x}{x^4-2x^2-1}\,dx$

my solution is $-\displaystyle\frac{1}{2(x^2-1)}$

The book only gives me these solutions where only 1 the correct one:

A) $\ln{\frac{x^2+2}{x^2+1}}$
B) $\ln{\frac{x^2+1}{x^2+2}}$
C) $\ln{|\frac{x^2-(\sqrt{2}+1)}{x^2+(\sqrt{2}-1)}|}$
D) $\frac{1}{4\sqrt{2}}\ln{|\frac{x^2-(\sqrt{2}+1)}{x^2+(\sqrt{2}-1)}|}$
E) $\ln{|\frac{x^2-1}{x^2+1}|}$
F) $\ln{|\frac{x^2+1}{x^2-1}|}$
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&#47;&#40;x&#94;4 - 2x&#94;2 - 1&#41; - Wolfram|Alpha (be sure to click on Show steps).

6. Ok, i'll post my method in a moment...

7. Antiderivatives are nessecarily the same up-to the addition of a constant. For, if we suppose that two functions $f, g$ both have the same derivative, then it follows that

$\displaystyle
(f - g)' = 0
$

and by the Mean Value Theorem, we know this means $f - g$ is constant.

8. Hello, Utherr!

Your algebra is off . . .

$\displaystyle\int\frac{x}{x^4-2x^2-1}\,dx$

$(A)\;\ln\left[\dfrac{x^2+2}{x^2+1}\right] + C \qquad\qquad\quad (B)\;\ln\left[\dfrac{x^2+1}{x^2+2}\right] + C$

$(C)\; \ln\left|\dfrac{x^2-(\sqrt{2}+1)}{x^2+(\sqrt{2}-1)}\right| + C \qquad
(D)\;\dfrac{1}{4\sqrt{2}}\ln\left|\dfrac{x^2-(\sqrt{2}+1)}{x^2+(\sqrt{2}-1)}\right| + C$

$(E)\;\ln\left|\dfrac{x^2-1}{x^2+1}\right| + C \qquad\qquad\quad\; (F)\;\ln\left|\dfrac{x^2+1}{x^2-1}\right|+C$

The denominator is: . $x^4 - 2x^2 - 1 \;=\;x^4 - 2x^2 + 1 - 2 \;=\;(x^2-1)^2 - 2$

The integral is: . $\displaystyle{\int\frac{x\,dx}{(x^2-1)^2 - 2} }$

Let: . $u \,=\,x^2-1\quad\Rightarrow\quad du \,=\,2x\,dx \quad\Rightarrow\quad x\,dx \,=\,\frac{1}{2}du$

. . Substitute: . $\displaystyle{\int\frac{\frac{1}{2}du}{u^2-2} \;=\;\tfrac{1}{2}\int\frac{du}{u^2- (\sqrt{2})^2} }$

Formula: . $\displaystyle{\int \frac{du}{u^2-a^2} \;=\;\frac{1}{2a}\ln\left|\frac{u-a}{u+a}\right| + C}$

. . We have: . $\displaystyle{\frac{1}{2}\cdot\frac{1}{2\sqrt{2}}\ ln\left|\frac{u-\sqrt{2}}{u+\sqrt{2}}\right| + C \;=\;\frac{1}{4\sqrt{2}}\ln\left|\frac{u-\sqrt{2}}{u+\sqrt{2}}\right| + C }$

Back-substitute: . $\displaystyle{\frac{1}{4\sqrt{2}}\ln\left|\frac{(x ^2-1) - \sqrt{2}}{(x^2-1) + \sqrt{2}}\right| + C }$

. . . . . . . . . . . $\displaystyle{=\;\frac{1}{4\sqrt{2}}\ln\left|\frac {x-(\sqrt{2}+1)}{x + (\sqrt{2}-1)}\right| + C } \quad\hdots\;\text{answer (D)}$

9. Hello again Utherr!

A function can have different-looking antiderivatives
. . but they are all equivalent.

Here is a classic example: . $\displaystyle{\int \sin x\cos x\,dx}$

$\displaystyle{[1]\;\int\sin x(\cos x\,dx) }$

Let $u = \sin x \quad\Rightarrow\quad du = \cos x\,dx$

Substitute: . $\displaystyle{\int u\,du \:=\:\tfrac{1}{2}u^2 + C}$

Back-substitute: . $\boxed{\tfrac{1}{2}\sin^2\!x + C}$

$\displaystyle{[2]\;\int \cos x(\sin x\,dx) }$

Let $u = \cos x \quad\Rightarrow\quad du = -\sin x\,dx \quad\Rightarrow\quad \sin x\,dx = -du$

Substitute: . $\displaystyle{\int u (-du) \;=\;-\int u\,du \;=\;-\tfrac{1}{2}u^2+C }$

Back-substitute: . $\boxed{-\tfrac{1}{2}\cos^2\!x + C}$

$\displaystyle{[3]\;\int\sin x\cos x\,dx \;=\;\tfrac{1}{2}\int 2\sin x\cos x\,dx$

. . . $\displaystyle{=\;\tfrac{1}{2}\int\sin2x\,dx \;=\;\boxed{-\tfrac{1}{4}\cos^2\!2x + C} }$

10. Originally Posted by Utherr
Ok, i'll post my method in a moment...

You stated originally that you wanted to integrate $\int \frac{xdx}{x^4- 2x^2- 1}$ but here you are integrating $\int \frac{xdx}{x^4- 2x^2+ 1}$.

To integrate $\int\frac{xdx}{x^4- 2x^2- 1}$, write it as $\int \frac{xdx}{x^2- 2x^2+ 1- 2}= \int\frac{xdx}{(x^2-1)^2- 2}$. Now let $y= x^2- 1$ so that dy= 2xdx and the integral becomes $\frac{1}{2}\int \frac{dy}{y^2- 2}= \frac{1}{2}\int \frac{dy}{(y- \sqrt{2})(y+ \sqrt{2})}$ which can be integrated by "partial fractions".

11. Thanks for the answers, i thought it was a + there not a - sign O.o