Mathematics

Evaluate:$$\displaystyle\int{\dfrac{dx}{{x}^{2}{\left({x}^{4}+1\right)}^{\frac{3}{4}}}}$$


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Subjective Medium Published on 17th 09, 2020
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Q1 Single Correct Medium
$$\displaystyle \int\frac{dx}{x\sqrt{1-x^{3}}}=$$
  • A. $$\dfrac{1}{3}\log|\displaystyle \frac{\sqrt{1-x^{2}}+1}{\sqrt{1-x^{2}}-1}|+c$$
  • B. $$\dfrac{1}{3}\log|\displaystyle \frac{1}{\sqrt{1-x^{3}}}|+c$$
  • C. $$\displaystyle \frac{1}{3}\log|1-x^{3}|+c$$
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1 Verified Answer | Published on 17th 09, 2020

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Q2 Single Correct Medium
$$\displaystyle \frac{x^{2}}{(x^{2}+a^{2})(x^{2}+b^{2})}=k \left [\displaystyle \frac{a^{2}}{x^{2}+a^{2}}-\displaystyle \frac{b^{2}}{x^{2}+b^{2}} \right]\Rightarrow k=$$
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Q3 Subjective Hard
Integrate :$$\displaystyle\int{{\left({\left(\dfrac{x+1}{x-1}\right)}^{2}+{\left(\dfrac{x-1}{x+1}\right)}^{2}-2\right)}^{\frac{1}{2}}dx}$$

Asked in: Mathematics - Integrals


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Q4 Single Correct Medium
Antiderivation of $$\dfrac {\sin^{2}x}{1+\sin^{2}x}$$ w.r.t is:
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  • B. $$x+\sqrt {2} arc\tan(\sqrt {2}\tan x)+C$$
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Q5 Single Correct Medium
$$\int\limits_{ - 5}^5 {\log \left( {\frac{{130 - {x^3}}}{{130 + {x^3}}}} \right)} dx$$ is equal to
  • A. $$\log \frac{{57}}{5}$$
  • B. $$2\int\limits_{ - 5}^5 {\log \left( {\frac{{130 - {x^3}}}{{130 + {x^3}}}} \right)} dx$$
  • C. $$-1$$
  • D. $$0$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

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