Question

Solve ${\int }_4^7\frac{1}{(x-3{)}^2}dx$

• A. $\frac{1}{27}$
• B. $\frac{3}{4}$
• C. $\frac{1}{9}$
• D. $\ln 4-\ln 3$
• E. $\ln 6-\ln 9$

Answer 1

The correct option is B $\frac{3}{4}$

We need to find ${\int }_4^7\frac{1}{{(x-3)}^2}dx$
Let us consider $x-3=t$

$\Rightarrow dx=dt$
$\Rightarrow {\int }_1^4\frac{1}{t^2}dt=-(\frac{1}{2}-\frac{1}{1})=1-\frac{1}{4}=\frac{3}{4}$