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

$du = 2dx$

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

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

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

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

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

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Subjective Medium Published on 17th 09, 2020
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