Definite Integration Question 7
Question 7 - 25 January - Shift 2
The integral $16 \int_1^{2} \frac{dx}{x^{3}(x^{2}+2)^{2}}$ is equal to
(1) $\frac{11}{6}+\log _e 4$
(2) $\frac{11}{12}+\log _e 4$
(3) $\frac{11}{12}-\log _e 4$
(4) $\frac{11}{6}-\log _e 4$
Show Answer
Answer: (4)
Solution:
Formula: Integration by substitution, Standard formulas for Indefinite Integration
$ \begin{aligned} & I=16 \int_1^{2} \frac{d x}{x^{3}(x^{2}+2)^{2}} \\ & =16 \int_1^{2} \frac{d x}{x^{3} x^{4}(1+\frac{2}{x^{2}})^{2}} \end{aligned} $
Substitute $1+\frac{2}{x^{2}}=t \Rightarrow \frac{-4}{x^{3}} dx=dt$
$ I=-4 \int_3^{\frac{3}{2}} \frac{d t}{(\frac{2}{t-1})^{2} t^{2}} $
$I=-4 \int_3^{\frac{3}{2}}(\frac{t-1}{2})^{2} \frac{dt}{t^{2}}$
$I=-\frac{4}{4} \int_3^{\frac{3}{2}}(1-\frac{2}{t}+\frac{1}{t^{2}}) d t$
$I=-1[t-2 \ln |t|-\frac{1}{t}]_3^{\frac{3}{2}}$
$I=-1[(\frac{3}{2}-2 \ln \frac{3}{2}-\frac{2}{3})-(3-2 \ln 3-\frac{1}{3})]$
$I=-1[2 \ell n 2-\frac{11}{6}]$
$I=\frac{11}{6}-\ln 4$
