Advertisement Remove all ads

# ∫ X 2 √ X − 1 D X - Mathematics

Advertisement Remove all ads
Advertisement Remove all ads
Advertisement Remove all ads
Sum
$\int\frac{x^2}{\sqrt{x - 1}} dx$
Advertisement Remove all ads

#### Solution

$\int\frac{x^2}{\sqrt{x - 1}}\text{ dx }$
$\text{Let x} - 1 = t^2$
$\Rightarrow x = t^2 + 1$
$\Rightarrow 1 = 2t \frac{dt}{dx}$
$\Rightarrow dx = \text{ 2t dt }$
$Now, \int\frac{x^2}{\sqrt{x - 1}}\text{ dx }$
$= \int\frac{\left( t^2 + 1 \right)^2}{t}\text{ 2t dt }$
$= 2\int\left( t^4 + 2 t^2 + 1 \right)dt$
$= 2\left[ \frac{t^{4 + 1}}{4 + 1} + \frac{2 t^{2 + 1}}{2 + 1} + t \right] + C$
$= 2\left[ \frac{t^5}{5} + \frac{2 t^3}{3} + t \right] + C$
$= 2\left[ \frac{3 t^5 + 10 t^3 + 15t}{15} \right] + C$
$= \frac{2}{15}t\left[ 3 t^4 + 10 t^2 + 15 \right] + C$
$= \frac{2}{15}\sqrt{x - 1} \left[ 3 \left( x - 1 \right)^2 + 10\left( x - 1 \right) + 15 \right] + C$
$= \frac{2}{15}\sqrt{x - 1} \left[ 3\left( x^2 - 2x + 1 \right) + 10x - 10 + 15 \right] + C$
$= \frac{2}{15}\sqrt{x - 1} \left[ 3 x^2 - 6x + 3 + 10x - 10 + 15 \right] + C$
$= \frac{2}{15}\sqrt{x - 1}\left[ 3 x^2 + 4x + 8 \right] + C$

Concept: Indefinite Integral Problems
Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 12 Maths
Chapter 19 Indefinite Integrals
Exercise 19.1 | Q 2 | Page 65

#### Video TutorialsVIEW ALL [1]

Advertisement Remove all ads
Share
Notifications

View all notifications

Forgot password?