Question

Evaluate the integral\(\int {\frac{{{{\rm{x}}^{\rm{2}}}{\rm{ + 8x - 3}}}}{{{{\rm{x}}^{\rm{3}}}{\rm{ + 3}}{{\rm{x}}^{\rm{2}}}}}{\rm{dx}}} \).

Short answer

The integral value of the given equation is\(\int {\frac{{{{\rm{x}}^{\rm{2}}}{\rm{ + 8x - 3}}}}{{{{\rm{x}}^{\rm{3}}}{\rm{ + 3}}{{\rm{x}}^{\rm{2}}}}}{\rm{dx = 3ln}}\left| {\rm{x}} \right|{\rm{ + }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{ - 2ln}}\left| {{\rm{x - 3}}} \right|{\rm{ + C}}} \).

Step-by-step solution

Expand the equation.

Decompose the partial fractions. Because is a repeating factor, you must write a fraction for each time it occurs\({\rm{(2t)}}\). The repeating factor is used as the numerator for each of those fractions\({\rm{x}}\). As a result, they're essentially constants with a numerator. would not be like.

\(\begin{aligned}{l}\frac{{{{\rm{x}}^{\rm{2}}}{\rm{ + 8x - 3}}}}{{{{\rm{x}}^{\rm{3}}}{\rm{ + 3}}{{\rm{x}}^{\rm{2}}}}}\\{\rm{ = }}\frac{{{{\rm{x}}^{\rm{2}}}{\rm{ + 8x - 3}}}}{{{{\rm{x}}^{\rm{2}}}{\rm{(x + 3)}}}}\\{\rm{ = }}\frac{{\rm{A}}}{{\rm{x}}}{\rm{ + }}\frac{{\rm{B}}}{{{{\rm{x}}^{\rm{2}}}}}{\rm{ + }}\frac{{\rm{C}}}{{{\rm{x + 3}}}}\end{aligned}\)

Multiply both sides of the equation by a factor of two\({{\rm{x}}^{\rm{2}}}{\rm{(x + 3)}}\)

\({{\rm{x}}^{\rm{2}}}{\rm{ + 8x - 3 = Ax(x + 3) + B(x + 3) + C}}{{\rm{x}}^{\rm{2}}}{\rm{ - - - - - - - - - - - - - - - (1)}}\)

If\({\rm{x = 0}}\)

\(\begin{aligned}{c}{\rm{0 + 0 - 3 = 0 + B(0 + 3) + 0}}\\{\rm{ - 3 = 3 B}}\\{\rm{B = - 1}}\end{aligned}\)

If\({\rm{x = - 3}}\)

\(\begin{aligned}{c}{\rm{9 - 24 - 3 = 0 + 0 + C(9)}}\\{\rm{ - 18 = 9 C}}\\{\rm{C = - 2}}\end{aligned}\)

Find the \({\rm{A}}\)value in equation\({\rm{(1)}}\).

Such as\({\rm{x = 1}}\)then solve for\({\rm{A}}\)

\(\begin{aligned}{c}{\rm{1 + 8 - 3 = A(1)(1 + 3) + - 1(1 + 3) - 2(1)}}\\{\rm{6 = 4A - 6}}\\{\rm{12 = 4A}}\\{\rm{A = 3}}\end{aligned}\)

Parts integration yields,

\(\begin{aligned}{c}\int {\frac{{{{\rm{x}}^{\rm{2}}}{\rm{ + 8x - 3}}}}{{{{\rm{x}}^{\rm{3}}}{\rm{ + 3}}{{\rm{x}}^{\rm{2}}}}}} {\rm{dx = }}\int {\left( {\frac{{\rm{3}}}{{\rm{x}}}{\rm{ + }}\frac{{{\rm{ - 1}}}}{{{{\rm{x}}^{\rm{2}}}}}{\rm{ + }}\frac{{{\rm{ - 2}}}}{{{\rm{x + 3}}}}} \right)} {\rm{dx}}\\{\rm{ = 3ln|x| - ( - 1)}}{{\rm{x}}^{{\rm{ - 1}}}}{\rm{ - 2ln|x + 3| + C}}\\{\rm{ = 3ln|x| + }}\frac{{\rm{1}}}{{\rm{x}}}{\rm{ - 2ln|x + 3| + C}}\end{aligned}\)

Therefore, the integral value of the given equation is \(\int {\frac{{{{\rm{x}}^{\rm{2}}}{\rm{ + 8x - 3}}}}{{{{\rm{x}}^{\rm{3}}}{\rm{ + 3}}{{\rm{x}}^{\rm{2}}}}}} {\rm{dx = 3ln|x| + }}\frac{{\rm{1}}}{{\rm{x}}}{\rm{ - 2ln|x + 3| + C}}\).