EXERCISE 7.5

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EXERCISE 7.5

9.

    \[\begin{aligned} & \int \frac{d x}{x^{2}-3 x-4} \\ & \frac{1}{x^{2}-3 x-4}=\frac{1}{x^{2}-4 x+x-4}=\frac{1}{x(x-4)+1(x-4)} \\ & =\frac{1}{(x-4)(x+1)} \end{aligned}\]

Partial fraction decomposition

    \[\frac{1}{(x-4)(x+1)}=\frac{A}{x-4}+\frac{B}{x+1} \quad \text { (1) }\]

Multiplying by (x-4)(x+1)

    \[1=A(x+1)+B(x-4) \quad \text { (2) }\]

Put x+1=0 \Rightarrow x=-1 into (2)

    \[\begin{aligned} & 1=A(-1+1)+B(-1-4) \\ & 1=-5 B \Rightarrow B=-\frac{1}{5} \end{aligned}\]

Put x-4=0 \Rightarrow x=4 in to (2)

    \[1=A(4+1) \Rightarrow A=\frac{1}{5}\]

Substituting A=\frac{1}{5} and B=-\frac{1}{5} into (1)

    \[\frac{1}{(x-4)(x+1)}=\frac{1}{5(x-4)}-\frac{1}{5(x+1)}\]

    \[\begin{aligned} \int \frac{1}{(x-4)(x+1)} d x & =\frac{1}{5} \int \frac{d x}{x-4}-\frac{1}{5} \int \frac{1}{x+1} d x \\ & =\frac{1}{5} \ln |x-4|-\frac{1}{5} \ln |x+1|+C \end{aligned}\]

10.

    \[\begin{aligned} & \int \frac{d x}{x^{2}-6 x-7} \\ & \begin{aligned} \frac{1}{x^{2}-6 x-7} & =\frac{1}{x^{2}-7 x+x-7}=\frac{1}{x(x-7)+1(x-7)} \\ & =\frac{1}{(x-7)(x+1)} \end{aligned} \end{aligned}\]

Partial fraction decomposition

    \[\frac{1}{(x-7)(x+1)}=\frac{A}{x-7}+\frac{B}{x+1}\]

Multiplying by (x-7)(x+1)

    \[1=A(x+1)+B(x-7)\]

Put x+1=0 \Rightarrow x=-1 into (2)

    \[\begin{aligned} & \text { Put } x+1=0 \Rightarrow \\ & 1=B(1-7) \Rightarrow \quad 1=-6 B \Rightarrow B=-\frac{1}{6} \\ & \quad x=7 \text { into (2) } \end{aligned}\]

Put x-7=0 \Rightarrow x=7 into (2)

    \[\begin{aligned} & \text { Put } x-7=0 \Rightarrow 1=8 A \Rightarrow A=\frac{1}{8} \\ & 1=A(7+1) \Rightarrow \begin{array}{l} \text { and } B=-\frac{1}{6} \text { into } \end{array} \end{aligned}\]

Substituting A=\frac{1}{8} and B=-\frac{1}{6} into (1)

    \[\begin{aligned} \frac{1}{(x-7)(x+1)} & =\frac{1}{8(x-7)}+\frac{1}{6(x+1)} \\ \int \frac{1}{(x-7)(x+1)} d x & =\frac{1}{8} \int \frac{1}{x-7} d x+\frac{1}{6} \int \frac{1}{x+1} d x \\ & =\frac{1}{8} \ln |x-7|+\frac{1}{6} \ln |x+1|+C \end{aligned}\]

11.

    \[\begin{aligned} \int \frac{11 x+17}{2 x^{2}+7 x-4}dx \end{aligned}\]

    \[\begin{aligned} \frac{11 x+17}{2 x^{2}+7 x-4} & =\frac{11 x+17}{2 x^{2}+8 x-x-4}=\frac{11 x+17}{2 x(x+4)-1(x+4)} \\ & =\frac{11 x+17}{(x+4)(2 x-1)} \end{aligned}\]

Partial fraction decomposition

    \[\frac{11 x+17}{(x+4)(2 x-1)}=\frac{A}{x+4}+\frac{B}{2 x-1} \quad \text { (1) }\]

Multiplying by (x+4)(2 x-1)

    \[\begin{array}{r} 11 x+17=A(2 x-1)+B(x+4) \quad \text { (2) }\\ \end{array}\]

put 2 x-1=0 \Rightarrow x=\frac{1}{2} \quad into (2)

    \[\begin{aligned} & 11\left(\frac{1}{2}\right)+17=B\left(\frac{1}{2}+4\right) \\ & \frac{11+34}{2}=B\left(\frac{1+8}{2}\right) \Rightarrow \frac{45}{2}=\frac{9}{2} B \\ & B=\frac{45}{9}=5 \end{aligned}\]

Put x+4=0 \Rightarrow x=-4 into (2)

    \[\begin{aligned} 11(-4)+17 & =A(2(-4)-1) \\ -44+17 & =A(-9) \Rightarrow 27=9 A \\ A & =\frac{27}{9}=3 \end{aligned}\]

Substituting A=3 and B=5 into (1)

    \[\begin{aligned} \frac{11 x+17}{(x+4)(2 x-1)} & =\frac{3}{x+4}+\frac{5}{2 x-1} \\ \int \frac{11 x+17}{(x+4)(2 x-1)} d x & =3 \int \frac{1}{x+4} d x+5 \int \frac{1}{2 x-1} d x \\ & =3 \int \frac{1}{x+4} d x+\frac{5}{2} \int \frac{2}{2 x-1} d x \\ & =3 \ln |x+4|+\frac{5}{2} \ln |2 x-1|+C \end{aligned}\]

12.

    \[\begin{aligned} & \int \frac{5 x-5}{3 x^{2}-8 x-3} d x \\ & \begin{aligned} \frac{5 x-5}{3 x^{2}-8 x-3} & =\frac{5 x-5}{3 x^{2}-9 x+x-3}=\frac{5 x-5}{3 x(x-3)+1(x-3)} \\ & =\frac{5 x-5}{(3 x+1)(x-3)} \end{aligned} \end{aligned}\]

Partial fraction decomposition

    \[\frac{5 x-5}{(3 x+1)(x-3)}=\frac{A}{3 x+1}+\frac{B}{x-3} \quad \text { (1) }\]

Multiplying by (3 x+1)(x-3)

    \[\begin{aligned} 5 x-5 & =A(x-3)+B(3 x+1) \quad \text { (2) } \\ \text { Put } x & =3 \text { in to (2) } \\ 5(3)-5 & =B(3(3)+1) \Rightarrow 15-5=10 B \\ 10 & =10 B \Rightarrow B=1 \end{aligned}\]

Put 3 x+1=0 \Rightarrow x=-\frac{1}{3} in to (2)

    \[\begin{gathered} 5\left(-\frac{1}{3}\right)-5=A\left(-\frac{1}{3}-3\right) \\ \frac{-5-15}{3}=A\left(\frac{-1-9}{3}\right) \\ \frac{-20}{3}=\frac{-10}{3} A \\ \Rightarrow A=2 \end{gathered}\]

substituting A=2 and B=1 into (1)

    \[\begin{aligned} \frac{5 x-5}{(3 x+1)(x-3} & =\frac{2}{3 x+1}+\frac{1}{x-3} \\ \int \frac{5 x-5}{(3 x+1)(x-3)} d x & =2 \int \frac{1}{3 x+1} d x+\int \frac{1}{x-3} d x \\ & =\frac{2}{3} \int \frac{3}{3 x+1} d x+\int \frac{1}{x-3} d x \\ & =\frac{2}{3} \ln |3 x+1|+\ln |x-3|+c \end{aligned}\]

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