How do you find #int(x^2-1/x^2+root3x)dx#?

Question

Charles Arocha · Accepted Answer

#1/3x^3+1/x+3/4x^(4/3)+c#

integrate each term using the #color(blue)"power rule for integration"#

#• int(ax^n)=a/(n+1)x^(n+1) ; n!=-1#

#rArrint(x^2-1/x^2+root(3)(x))dx#

#=int(x^2-x^-2+x^(1/3))larr" in exponent form"#

#=1/3x^3+1/x+3/4x^(4/3)+c#

where c is the integration constant.

Christopher Agresta · Answer

#=x^3/3+1/x+3/4x^(4/3)+C#

#int(x^2-1/x^2+root(3)(x)) dx#

By revising,

#=int(x^2-x^(-2)+x^(1/3)) dx#

By the Power Rule for integration, which states that #intx^ndx=x^(n+1)/(n+1)+C#:

#=x^3/3-x^(-1)/(-1)+x^(4/3)/(4/3)+C#

By tidying up,

#=x^3/3+1/x+3/4x^(4/3)+C#

I hope this made sense.

Cameron Alexander · Answer

undefined To find $ \int \frac{x^2 - 1}{x^2 + \sqrt{3}x} \, dx $:

1. Perform polynomial long division to rewrite the integrand in the form $ Q(x) + \frac{R(x)}{x^2 + \sqrt{3}x} $, where $ Q(x) $ is a polynomial and $ R(x) $ is the remainder.
   $ \frac{x^2 - 1}{x^2 + \sqrt{3}x} = Q(x) + \frac{R(x)}{x^2 + \sqrt{3}x} $

2. After performing the polynomial long division, integrate the rational function $ \frac{R(x)}{x^2 + \sqrt{3}x} $ using partial fraction decomposition.

3. Once you have the integral in terms of partial fractions, integrate each term separately.

4. Combine the integrals obtained in step 3 with the integral of the polynomial $ Q(x) $ obtained from polynomial long division.

5. Evaluate the definite integral if necessary by substituting the limits of integration.

6. Simplify the expression if possible.