Take the highest power of x out of the bracket or root, then substitute t = 1 + 1/x^n

1. A.
   $-\frac{(x^4+1)^{1/4}}{x^4} + c$
2. B.
   $\frac{(x^4+1)^{1/4}}{x^4} + c$
3. C.
   $\left(1+\frac{1}{x^4}\right)^{1/4} + c$
4. D.
   $-\left(1+\frac{1}{x^4}\right)^{1/4} + c$

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1. A.
   $-\left(x^4+1\right)^{1/4}+c$
2. B.
   $-\frac{(x^4+1)^{1/4}}{x}+c$
3. C.
   $-\frac{(x^4+1)^{1/4}}{x}+c$ (same but negative)
4. D.
   $-\left(x^4+1\right)^{1/4}+c$

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1. A.
   $-\left(\frac{x^4+1}{x^4}\right)^{1/4}+c$
2. B.
   $\left(x^4+1\right)^{1/4}+c$
3. C.
   $-\left(x^4+1\right)^{1/4}+c$
4. D.
   $-\frac{(x^4+1)^{1/4}}{x}+c$

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1. A.
   $-\frac{(x^4+1)^{1/4}}{x}+c$
2. B.
   $\frac{(x^4-1)^{1/4}}{x}+c$
3. C.
   $-\frac{(x^4-1)^{1/4}}{x}+c$
4. D.
   $-\frac{(x^4+1)^{1/4}}{x}+c$

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1. A.
   $\sqrt[2]{2-\frac{2}{x^{2}}+\frac{1}{x^{4}}}+c$
2. B.
   $\frac{1}{2}\sqrt{2-\frac{2}{x^{2}}+\frac{1}{x^{4}}}+c$
3. C.
   $\sqrt[2]{2+\frac{2}{x^{2}}+\frac{1}{x^{4}}}+c$
4. D.
   $\sqrt[2]{2-\frac{2}{x^{2}}-\frac{1}{x^{4}}}+c$

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1. A.
   $-\!\left(\dfrac{x^4+1}{x^4}\right)^{1/4}+c$
2. B.
   $\left(\dfrac{x^4+1}{1}\right)^{1/4}+c$
3. C.
   $-\!\left(\dfrac{x^4+1}{1}\right)^{1/4}+c$
4. D.
   $-\!\left(\dfrac{x^4+1}{x^4}\right)^{1/4}+c$

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1. A.
   $\frac{1}{9x^4}$
2. B.
   $-\frac{1}{3x^3}$
3. C.
   $-\frac{1}{27x^9}$
4. D.
   $\frac{1}{27x^6}$

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1. A.
   $-\dfrac{1}{16}$
2. B.
   $-\dfrac{25}{16}$
3. C.
   $\dfrac{25}{4}$
4. D.
   $-\dfrac{25}{4}$

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