int (x^3 - 4x^2 - 4x + 20)/(x^2 - 5) dx Find the indefinite integral.

int (x^3-4x^2-4x+20)/(x^2-5)dx

To solve, divide the numerator by the denominator (see attached figure).

= int (x - 4 + x/(x^2-5))dx

= int xdx - int4dx + int x/(x^2-5)dx

For the first integral, apply the formula int x^ndx = x^(n+1)/(n+1)+ C.

For the second integral, apply the formula int adx = ax + C .

= x^2/2 - 4x +C + int x/(x^2-5)dx

For the third integral, apply u-substitution method.

Let

u = x^2-5

Differentiate u.

du=2x dx

(du)/2 =xdx

Plug-in them to the third integral.

=x^2/2 - 4x + C + int 1/(x^2-5)*xdx

=x^2/2 - 4x + C + int 1/u *(du)/2

= x^2/2 - 4x + C + 1/2int 1/u du

Then, apply the formula int 1/xdx = ln|x| +  C.

=x^2/2-4x + 1/2ln|u| + C

And, substitute back  u = x^2-5 .

=x^2/2 - 4x +1/2ln|x^2-5|+C

Therefore,  int (x^3-4x^2-4x + 20)/(x^2-5)dx = x^2/2 - 4x + 1/2ln|x^2-5|+C .

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