First slide

Methods of integration

Question

$Anti derivative of \frac{1}{\sqrt{(x + a) (x + b)}} d x is equal to$

Moderate

A

$\log |x + \frac{a + b}{2} + \sqrt{x^{2} + (a + b) x - a b}| + k$
B

$\log |x + \frac{a + b}{2} + \sqrt{x^{2} + (a + b) x}| + k$
C

$\log |x + \frac{a + b}{2} + \sqrt{x^{2} + (a + b) x + a b}| + k$
D

$\log |x - \frac{a + b}{2} + \sqrt{x^{2} + (a + b) x + a b}| + k$

Solution

$I = \int \frac{d x}{\sqrt{(x + a) (x + b)}}$
$\begin{array}{l} I = \int \frac{d x}{\sqrt{x^{2} + (a + b) x + a b}} \\ I = \int \frac{d x}{\sqrt{x^{2} + 2 \frac{(a + b)}{2} x + a b + {(\frac{a + b}{2})}^{2} - {(\frac{a + b}{2})}^{2}}} \\ I = \int \frac{d x}{\sqrt{{[x + \frac{a + b}{2}]}^{2} + \frac{4 a b - (a + b)^{2}}{4}}} \\ I = \int \frac{d x}{\sqrt{{[x + \frac{a + b}{2}]}^{2} - {[\frac{a - b}{2}]}^{2}}} \\ I = \log |x + (\frac{a + b}{2}) + \sqrt{x^{2} + (a + b) x + a b}| + k \end{array}$

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