First slide

Methods of integration

Question

If $\int \frac{d x}{x^{3} (x - 1)^{1 / 2}} = \frac{\sqrt{x - 1} (3 x + 2)}{4 x^{2}} +$
$K \tan^{- 1} \sqrt{x - 1} + C$ then the value of $K$ is

Moderate

A

1/2
B

1
C

1/4
D

3/4

Solution

Put $x - 1 = t^{2}$ so that $d x = 2 t d t$ and
$\int \frac{d x}{x^{3} (x - 1)^{1 / 2}} = \int \frac{2 t d t}{{(t^{2} + 1)}^{3} t} = 2 \int \frac{d t}{{(t^{2} + 1)}^{3}}$
Now we use formula of $\int \frac{d y}{{(y^{2} + k^{2})}^{n}}$ to write
$\int \frac{d t}{{(t^{2} + 1)}^{3}} = \frac{t}{4 {(t^{2} + 1)}^{2}} + \frac{3}{4} \int \frac{d t}{{(t^{2} + 1)}^{2}}$ and
$\begin{array}{l} \int \frac{d t}{{(t^{2} + 1)}^{2}} = \frac{t}{2 (t^{2} + 1)} + \frac{1}{2} \int \frac{d t}{t^{2} + 1} \\ = \frac{t}{2 (t^{2} + 1)} + \frac{1}{2} \tan^{- 1} t \end{array}$
Hence $\int \frac{d x}{x^{3} (x - 1)^{1 / 2}}$
$\begin{array}{l} = \frac{t}{2 {(t^{2} + 1)}^{2}} + \frac{3 t}{4 (t^{2} + 1)} + \frac{3}{4} \tan^{- 1} t + C \\ = \frac{1}{4 {(t^{2} + 1)}^{2}} (2 t + 3 t^{3} + 3 t) + \frac{3}{4} \tan^{- 1} t + C \\ = \frac{\sqrt{x - 1} (3 x + 2)}{4 x^{2}} + \frac{3}{4} \tan^{- 1} \sqrt{x - 1} + C . \end{array}$

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