Answer to Question #101487 in Calculus for Mal

Question

Integrate using the fractional fracture development (point b). For a and c, first divide by the division angle.

a.) ∫3x2+2x+2/x-3 dx

b.) ∫3/x2-1 dx

c.) ∫2x2+x/x+2 dx

Answer 1

Solution.

a. We write the integral in the form

$$\intop \frac {3x^2+2x+2}{x-3}dx=\intop \frac {3x(x-3)+11x+2}{x-3}dx=$$

$$=\intop \frac {3x(x-3)+11(x-3)+35}{x-3}dx= \intop (3x+11+\frac{35} {x-3})dx$$

Using the tabular values of the antiderivative we get

$$\intop \frac {3x^2+2x+2}{x-3}dx= \frac{3x^2}{2}+11x+35 ln |x-3| +C$$

where C is constant.

b. Represent the fraction 3/x2-1 as the sum of two fractions

$$\frac {3} {x^2-1}=\frac{A}{x-1} + \frac {B}{x+1}= \frac{Ax+A+Bx-B}{x^2-1}$$

Comparing the coefficients near the powers of x we get the system of equations

$$\begin{cases} A+B=0 \\ A-B=3 \end{cases} \implies \begin{cases} A=\frac{3}{2} \\ B=-\frac {3}{2} \end{cases}$$

We write the integral in the form

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

Using the tabular values of the antiderivative we get

$$\intop \frac {3}{x^2-1}dx=\frac{3}{2} ln|x-1|-\frac{3}{2} ln|x+1|+C=\frac{3}{2} ln \frac {|x-1|}{|x+1|}+C$$

where C is constant.

c. We write the integral in the form

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

$$\intop \frac {2x(x+2)-3(x+2) +6}{x+2}dx=\intop (2x-3+\frac {6}{x+2})dx$$

Using the tabular values of the antiderivative we get

$$\intop \frac {2x^2+x}{x+2}dx=x^2-3x+6ln|x+2|+C$$

where C is constant.

Answer. a.

$$\intop \frac {3x^2+2x+2}{x-3}dx= \frac{3x^2}{2}+11x+35 ln |x-3| +C$$

where C is constant.

b.

$$\intop \frac {3}{x^2-1}dx=\frac{3}{2} ln \frac {|x-1|}{|x+1|}+C$$

where C is constant.

c.

$$\intop \frac {2x^2+x}{x+2}dx=x^2-3x+6ln|x+2|+C$$

where C is constant.