Name: Second Fundamental Theorem of Calculus
Uploaded: 2019-07-19
Duration: 5 min 22 s
Description: In this video, we will learn about the evaluating the definite integral by using the second fundamnetal theorem of calculus in details.

Question

Evaluate $\int \frac{x + 2}{\sqrt{(x - 2) (x - 3)}} d x$

Accepted Answer

Let $I = \int \frac{x + 2}{\sqrt{(x - 2) (x - 3)}} d x$

$I = \int \frac{x + 2}{\sqrt{x^{2} - 5 x + 6}} d x$

$I = \int \frac{(\frac{1}{2} (2 x - 5) + \frac{9}{2})}{\sqrt{x^{2} - 5 x + 6}} d x$

$I = \frac{1}{2} \int \frac{2 x - 5}{\sqrt{x^{2} - 5 x + 6}} d x + \frac{9}{2} \int \frac{d x}{\sqrt{x^{2} - 5 x + {(\frac{5}{2})}^{2} - {(\frac{5}{2})}^{2} + 6}}$

$I = \frac{1}{2} \int \frac{2 x - 5}{\sqrt{x^{2} - 5 x + 6}} d x + \frac{9}{2} \int \frac{d x}{\sqrt{{(x - \frac{5}{2})}^{2} - \frac{25}{4} + 6}}$

$I = \frac{1}{2} \int \frac{2 x - 5}{\sqrt{x^{2} - 5 x + 6}} d x + \frac{9}{2} \int \frac{d x}{\sqrt{{(x - \frac{5}{2})}^{2} - \frac{1}{4}}}$

$I = \frac{1}{2} \int \frac{2 x - 5}{\sqrt{x^{2} - 5 x + 6}} d x + \frac{9}{2} \int \frac{d x}{\sqrt{{(x - \frac{5}{2})}^{2} - {(\frac{1}{2})}^{2}}}$

Using: $\int \frac{d x}{\sqrt{x^{2} - a^{2}}} = \log |x + \sqrt{x^{2} - a^{2}}| + c$ and $\int \frac{f^{'} (x)}{\sqrt{f (x)}} d x = 2 \sqrt{f (x)} + c$

Hence, $I = \sqrt{x^{2} - 5 x + 6} + \frac{9}{2} \log | (x - \frac{5}{2}) + \sqrt{x^{2} - 5 x + 6} | + c$ .

Evaluate $\int \frac{x + 2}{\sqrt{(x - 2) (x - 3)}} d x$

Important Questions on Integrals

Learn and Prepare for any exam you want !

Evaluate ∫x+2(x-2)(x-3) dx

Important Questions on Integrals

Learn and Prepare for any exam you want !

Evaluate $\int \frac{x + 2}{\sqrt{(x - 2) (x - 3)}} d x$