evaluate each indefinite integral.\n1) $int\frac{1}{sqrt{16 - x^{2}}}dx$\n2) $int\frac{1}{4 + x^{2}}dx$\n3)…

evaluate each indefinite integral.\n1) $int\frac{1}{sqrt{16 - x^{2}}}dx$\n2) $int\frac{1}{4 + x^{2}}dx$\n3) $int\frac{1}{xsqrt{x^{2}-1}}dx$\n4) $int\frac{1}{16 + x^{2}}dx$\n5) $int\frac{1}{xsqrt{x^{2}-4}}dx$\n6) $int\frac{1}{sqrt{25 - x^{2}}}dx$\n7) $int\frac{1}{xsqrt{x^{2}-81}}dx$\n8) $int\frac{1}{4 + x^{2}}dx$

evaluate each indefinite integral.\n1) $int\frac{1}{sqrt{16 - x^{2}}}dx$\n2) $int\frac{1}{4 + x^{2}}dx$\n3) $int\frac{1}{xsqrt{x^{2}-1}}dx$\n4) $int\frac{1}{16 + x^{2}}dx$\n5) $int\frac{1}{xsqrt{x^{2}-4}}dx$\n6) $int\frac{1}{sqrt{25 - x^{2}}}dx$\n7) $int\frac{1}{xsqrt{x^{2}-81}}dx$\n8) $int\frac{1}{4 + x^{2}}dx$

Answer

  1. For (\int\frac{1}{\sqrt{16 - x^{2}}}dx):
    • Explanation:
      • Recall the integral formula (\int\frac{1}{\sqrt{a^{2}-x^{2}}}dx=\arcsin(\frac{x}{a})+C). Here (a = 4) since (a^{2}=16).
      • So (\int\frac{1}{\sqrt{16 - x^{2}}}dx=\arcsin(\frac{x}{4})+C).
  2. For (\int\frac{1}{4 + x^{2}}dx):
    • Explanation:
      • Recall the integral formula (\int\frac{1}{a^{2}+x^{2}}dx=\frac{1}{a}\arctan(\frac{x}{a})+C). Here (a = 2) since (a^{2}=4).
      • So (\int\frac{1}{4 + x^{2}}dx=\frac{1}{2}\arctan(\frac{x}{2})+C).
  3. For (\int\frac{1}{x\sqrt{x^{2}-1}}dx):
    • Explanation:
      • Recall the integral formula (\int\frac{1}{x\sqrt{x^{2}-1}}dx=\text{arcsec}(|x|)+C).
      • So (\int\frac{1}{x\sqrt{x^{2}-1}}dx=\text{arcsec}(|x|)+C).
  4. For (\int\frac{1}{16 + x^{2}}dx):
    • Explanation:
      • Using the formula (\int\frac{1}{a^{2}+x^{2}}dx=\frac{1}{a}\arctan(\frac{x}{a})+C), with (a = 4) (since (a^{2}=16)).
      • So (\int\frac{1}{16 + x^{2}}dx=\frac{1}{4}\arctan(\frac{x}{4})+C).
  5. For (\int\frac{1}{x\sqrt{x^{2}-4}}dx):
    • Explanation:
      • Using the formula (\int\frac{1}{x\sqrt{x^{2}-a^{2}}}dx=\frac{1}{a}\text{arcsec}(\frac{|x|}{a})+C), with (a = 2) (since (a^{2}=4)).
      • So (\int\frac{1}{x\sqrt{x^{2}-4}}dx=\frac{1}{2}\text{arcsec}(\frac{|x|}{2})+C).
  6. For (\int\frac{1}{\sqrt{25 - x^{2}}}dx):
    • Explanation:
      • Using the formula (\int\frac{1}{\sqrt{a^{2}-x^{2}}}dx=\arcsin(\frac{x}{a})+C), with (a = 5) (since (a^{2}=25)).
      • So (\int\frac{1}{\sqrt{25 - x^{2}}}dx=\arcsin(\frac{x}{5})+C).
  7. For (\int\frac{1}{x\sqrt{x^{2}-81}}dx):
    • Explanation:
      • Using the formula (\int\frac{1}{x\sqrt{x^{2}-a^{2}}}dx=\frac{1}{a}\text{arcsec}(\frac{|x|}{a})+C), with (a = 9) (since (a^{2}=81)).
      • So (\int\frac{1}{x\sqrt{x^{2}-81}}dx=\frac{1}{9}\text{arcsec}(\frac{|x|}{9})+C).
  8. For (\int\frac{1}{4 + x^{2}}dx):
    • Explanation:
      • Using the formula (\int\frac{1}{a^{2}+x^{2}}dx=\frac{1}{a}\arctan(\frac{x}{a})+C), with (a = 2) (since (a^{2}=4)).
      • So (\int\frac{1}{4 + x^{2}}dx=\frac{1}{2}\arctan(\frac{x}{2})+C).

Answer:

  1. (\arcsin(\frac{x}{4})+C)
  2. (\frac{1}{2}\arctan(\frac{x}{2})+C)
  3. (\text{arcsec}(|x|)+C)
  4. (\frac{1}{4}\arctan(\frac{x}{4})+C)
  5. (\frac{1}{2}\text{arcsec}(\frac{|x|}{2})+C)
  6. (\arcsin(\frac{x}{5})+C)
  7. (\frac{1}{9}\text{arcsec}(\frac{|x|}{9})+C)
  8. (\frac{1}{2}\arctan(\frac{x}{2})+C)