# Thread: Need help with basic calculus

1. ## Need help with basic calculus

How can I integrate the following given?

int[x^2(1-x^3)/(x^2-1)]dx ans. -1/4x^4-1/2x^2+x-ln(x+1)+c

int[cos[1/2x]dx/3/4-2/3 sin[1/2x] ans. -3ln(3/4-2/3sin[1/2x]+c

int=integral symbol

Thanks

2. Hi honestliar
1.
$\int \frac{x^2(1-x^3)}{x^2-1}dx$ ; the numerator divided by denominator will be :

$=\int(-x^3 - x + 1 + \frac{1-x}{x^2-1})dx$

Solve $\int \frac{1-x}{x^2-1}dx$ using partial fraction

2.
hint : $\frac{d}{dx}sec (x) = tan(x) sex(x)$

3. Originally Posted by honestliar
How can I integrate the following given?

int[x^2(1-x^3)/(x^2-1)]dx

int[cos[1/2x]dx/3/4-2/3 sin[1/2x]

int=integral symbol

Thanks
$\int {\frac{{{x^2}\left( {1 - {x^3}} \right)}}{{{x^2} - 1}}dx} .$

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

$= - {x^2} - \frac{{{x^4}}}{{1 + x}} = - {x^2} + \frac{{1 - {x^4} - 1}}{{1 + x}} = - {x^2} + \frac{{1 - {x^4}}}{{1 + x}} - \frac{1}
{{1 + x}} =$

$- {x^2} + \frac{{\left( {1 - {x^2}} \right)\left( {1 + {x^2}} \right)}}{{1 + x}} - \frac{1}{{1 + x}} = - {x^2} + \frac{{\left( {1 - x} \right)\left( {1 + x} \right)\left( {1 + {x^2}} \right)}}{{1 + x}} - \frac{1}{{1 + x}} =$

$= - {x^2} + \left( {1 - x} \right)\left( {1 + {x^2}} \right) - \frac{1}
{{1 + x}} = 1 - x - {x^3} - \frac{1}{{1 + x}}.$

$\int {\frac{{{x^2}\left( {1 - {x^3}} \right)}}{{{x^2} - 1}}dx} = \int {\left( {1 - x - {x^3} - \frac{1}{{1 + x}}} \right)dx} .$