Mathematics

$$2 \int \dfrac{tdt}{t^4+1}=\dfrac{1}{\sqrt{m}}tan^{-1}{t^2}$$.Find the value of $$m$$


ANSWER

1


SOLUTION
$$2\int { \dfrac { t\;dt }{ { t }^{ 4 }+1 }  } $$
Let $$t^2=M$$
$$\Rightarrow 2+dt=dM$$       ( On differentiating w.r.t $$t$$ )
$$\Rightarrow \dfrac { 2 }{ 2 } \int { \dfrac { dM }{ { M }^{ 2 }+1 } =\tan ^{ -1 }{ M } +C } $$
                              $$=\tan^{-1}\left(t^2\right) +C.$$
Hence, the answer is $$\tan^{-1}\left(t^2\right) +C.$$

View Full Answer

Its FREE, you're just one step away


One Word Medium Published on 17th 09, 2020
Next Question
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84
Enroll Now For FREE

Realted Questions

Q1 Single Correct Hard
Evaluate: $$\displaystyle \int { x\log { \left( 1+x \right) dx }  } $$
  • A. $$\displaystyle \frac { { x }^{ 2 } }{ 2 } \ln { \left( x+1 \right)  } +\frac { 1 }{ 2 } \left[ \frac { { x }^{ 2 } }{ 2 } -x+\ln { \left| x+1 \right|  }  \right] +c$$
  • B. $$\displaystyle \frac { { x }^{ 2 } }{ 2 } \ln { \left( x+1 \right)  } +\frac { 1 }{ 2 } \left[ \frac { { x }^{ 2 } }{ 2 } +x+\ln { \left| x+1 \right|  }  \right] +c$$
  • C. none of these
  • D. $$\displaystyle \frac { { x }^{ 2 } }{ 2 } \ln { \left( x+1 \right)  } -\frac { 1 }{ 2 } \left[ \frac { { x }^{ 2 } }{ 2 } -x+\ln { \left| x+1 \right|  }  \right] +c$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q2 Subjective Medium
Evaluate $$\int _{ 0 }^{ 3 }{ \left( 2{ x }^{ 2 }+3x+5 \right) dx }$$ as limit of a sum.

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q3 Subjective Medium
Solve $$\displaystyle\int{\dfrac{\left( 4{{x}^{3}}+5{{x}^{4}} \right)}{{{\left( {{x}^{5}}+{{x}^{4}}+6 \right)}^{2}}}}dx$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q4 Subjective Hard
Integrate $$\displaystyle \int_{0}^{\frac {\pi}{4}} \log (1 +\tan x)dx$$.

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q5 Subjective Easy
Evaluate:
$$ \int_{}^{} {\frac{{ - 1}}{{\sqrt {1 - {x^2}} }}dx} $$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer