EXERCISE 7.5

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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