Mathematics

Evaluate:

$$ \displaystyle \int {\dfrac{{1 - {x^3}}}{{\left( {1 - 2x} \right)}}} \,dx$$


SOLUTION

$$\displaystyle \int {{{1 - {x^3}} \over {\left( {1 - 2x} \right)}}dx} $$


$$ = \displaystyle \int {{{{x^3} - 1} \over {2x - 1}}dx} $$


Let $$u = 2x - 1 \Rightarrow {{u + 1} \over 2} = x$$


$$du = 2dx$$


$${x^3} = {\left( {{{4 + 1} \over 2}} \right)^3}$$


$$ = {1 \over {16}}\displaystyle \int {{{\left( {4 - 1} \right)\left( {{4^2} + 4u + 7} \right)} \over 4}du} $$


$$ = {1 \over {16}}\displaystyle \int {\left( {{4^2} + 3u - {7 \over u} + 3} \right)du} $$


$$ = {1 \over {16}}\left[ {{{{u^3}} \over 3} + {{3{u^2}} \over 2} - 7lu\left( 4 \right) + 3u} \right]$$


$$ = {{{{\left( {2x - 1} \right)}^3}} \over {48}} + {{3{{\left( {2x - 1} \right)}^2}} \over {32}} - 7{\mathop{\rm l}\nolimits} u\left( {2x - 1} \right) + 3\left( {2x - 1} \right) + C$$

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 109
Enroll Now For FREE

Realted Questions

Q1 One Word Medium
$$2 \int \dfrac{tdt}{t^4+1}=\dfrac{1}{\sqrt{m}}tan^{-1}{t^2}$$.Find the value of $$m$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q2 Subjective Medium
$$ \int $$ $$\dfrac {dx}{(x^{2}+1)(x+3)}$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q3 Single Correct Medium
If $$f(x)=\int_{0}^{x}t (sinx - sint)dt$$  then:
  • A. $$f'''(x)+f''(x)-f'(x)=cosx$$
  • B. $$f'''(x)-f''(x)=cosx-2x sin$$
  • C. $$f'''(x)+f''(x)=sinx$$
  • D. $$f'''(x)+f'(x)= cosx - 2x sinx$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q4 Subjective Medium
Evaluate: $$\displaystyle\int  \dfrac{ dx} {{{\cos }^2}x{{\left( {1 - \tan x} \right)}^2}} $$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q5 Single Correct Medium
$$\displaystyle \int\frac{dt}{(6t-1)}$$ is equal to:
  • A. $$\ln(6t-1)+C$$
  • B. $$-\dfrac{1}{6}\ln(6t-1)+C$$
  • C. None of these
  • D. $$\dfrac{1}{6} \ln(6t-1) +C$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer