Mathematics

If $$\displaystyle \int {\left  (\dfrac { x-1 }{ { x }^{ 2 } } \right ){ e }^{ x }dx=f\left( x \right) { e }^{ x }+c } $$, then write the value of f(x).


SOLUTION
$$\displaystyle\int \dfrac{x-1}{x^2}\times e^{x} d x=\int e^{x}\bigg(\dfrac{1}{x}-\dfrac{1}{x^2}\bigg) dx$$

This is in the form of $$\displaystyle\int e^{x}(g(x)+g'(x)) d x=e^{x}g(x)+C$$

Here $$g(x)=\dfrac{1}{x}$$ and $$g'(x)=-\dfrac{1}{x^2}$$

$$\implies \displaystyle\int (\dfrac{x-1}{x^2})e^{x} d x=\dfrac{e^{x}}{x}+C$$

$$\implies f(x)=\dfrac{1}{x}$$
View Full Answer

Its FREE, you're just one step away


Subjective 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 Medium
$$\displaystyle \int\frac{x+\sin x}{1+\cos x}dx=$$
  • A. $$xcot \displaystyle \frac{x}{2}+c$$
  • B. $$xsin \displaystyle \frac{x}{2}+c$$
  • C. $$xcos \displaystyle \frac{x}{2}+c$$
  • D. $$xtan \displaystyle \frac{x}{2}+c$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q2 Single Correct Medium
$$\int {{e^{3{{\log }_e}x}}.{{\left( {{x^4} + 1} \right)}^{ - 1}}dx = \_\_\_\_\_\_\_\_\_ + C.} $$
  • A. $$\log \left( {{x^4} + 1} \right)$$
  • B. $$-\log \left( {{x^4} + 1} \right)$$
  • C. $$\frac{{ - 3}}{{{{\left( {{x^4} + 1} \right)}^3}}}$$
  • D. $$\frac{1}{4}\log \left( {{x^4} + 1} \right)$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q3 Single Correct Medium
The value of $$\displaystyle\int { \cfrac { \sin { x } +\cos { x }  }{ 3+\sin { 2x }  }  } dx$$ is
  • A. $$\cfrac { 1 }{ 2 } \log { \left( \cfrac { 2+\sin { x } }{ 2-\sin { x } } \right) +C } $$
  • B. $$\cfrac { 1 }{ 4 } \log { \left( \cfrac { 1+\sin { x } }{ 1-\sin { x } } \right) +C } $$
  • C. None of the above
  • D. $$\cfrac { 1 }{ 4 } \log { \left( \cfrac { 2-\sin { x } +\cos { x } }{ 2+\sin { x } -\cos { x } } \right) } +C$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q4 Subjective Hard
Evaluate: $$\int ^{1}_{0}x (\tan^{-1}x)^{2}dx$$

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