English

∫ 6 X − 5 √ 3 X 2 − 5 X + 1 D X - Mathematics

Advertisements
Advertisements

Question

\[\int\frac{6x - 5}{\sqrt{3 x^2 - 5x + 1}} \text{ dx }\]
Sum
Advertisements

Solution

\[\text{ Let I }= \int\frac{6x - 5}{\sqrt{3 x^2 - 5x + 1}}dx\]
\[\text{Putting}\  3 x^2 - 5x + 1 = t\]
\[ \Rightarrow \left( 6x - 5 \right) dx = dt\]
\[\text{ Then }, \]
\[I = \int\frac{dt}{\sqrt{t}}\]
\[ = 2\sqrt{t} + C\]
\[ = 2\sqrt{3 x^2 - 5x + 1} + C\]

shaalaa.com
  Is there an error in this question or solution?
Chapter 19: Indefinite Integrals - Exercise 19.21 [Page 110]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 19 Indefinite Integrals
Exercise 19.21 | Q 4 | Page 110

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

\[\int\frac{x^5 + x^{- 2} + 2}{x^2} dx\]

\[\int\left( \sec^2  x + {cosec}^2  x \right)  dx\]

\[\int\frac{5 \cos^3 x + 6 \sin^3 x}{2 \sin^2 x \cos^2 x} dx\]

If f' (x) = 8x3 − 2xf(2) = 8, find f(x)


\[\int\frac{2x + 3}{\left( x - 1 \right)^2} dx\]

` ∫   cos  3x   cos  4x` dx  

\[\int\sqrt{\frac{1 + \cos 2x}{1 - \cos 2x}} dx\]

\[\int\sqrt{\frac{1 - \sin 2x}{1 + \sin 2x}} dx\]

` ∫  {sin 2x} /{a cos^2  x  + b sin^2  x }  ` dx 


\[\int\frac{\cos x - \sin x}{1 + \sin 2x} dx\]

\[\int\frac{\left( x + 1 \right) e^x}{\sin^2 \left( \text{x e}^x \right)} dx\]

` ∫    x   {tan^{- 1} x^2}/{1 + x^4} dx`

\[\int\frac{2x - 1}{\left( x - 1 \right)^2} dx\]

\[\int \cot^5 \text{ x } {cosec}^4 x\text{ dx }\]

\[\int \sin^3 x \cos^6 x \text{ dx }\]

\[\int\frac{e^x}{1 + e^{2x}} dx\]

\[\int\frac{dx}{e^x + e^{- x}}\]

\[\int\frac{x}{x^4 + 2 x^2 + 3} dx\]

\[\int\frac{1}{\sqrt{5 x^2 - 2x}} dx\]

\[\int\frac{1}{x^{2/3} \sqrt{x^{2/3} - 4}} dx\]

\[\int\frac{x + 1}{x^2 + x + 3} dx\]

\[\int\frac{2x + 1}{\sqrt{x^2 + 2x - 1}}\text{  dx }\]

\[\int\frac{x + 1}{\sqrt{4 + 5x - x^2}} \text{ dx }\]

\[\int x^2 \text{ cos x dx }\]

\[\int\left( x + 1 \right) \text{ e}^x \text{ log } \left( x e^x \right) dx\]

\[\int\left( \tan^{- 1} x^2 \right) x\ dx\]

\[\int\frac{x^2 + x - 1}{x^2 + x - 6} dx\]

\[\int\frac{5 x^2 - 1}{x \left( x - 1 \right) \left( x + 1 \right)} dx\]

\[\int\frac{dx}{\left( x^2 + 1 \right) \left( x^2 + 4 \right)}\]

\[\int\frac{3x + 5}{x^3 - x^2 - x + 1} dx\]

\[\int\frac{x^2 + 1}{x^4 + x^2 + 1} \text{  dx }\]

\[\int\frac{x^2 + 1}{x^4 + 7 x^2 + 1} 2 \text{ dx }\]

\[\int\frac{x}{\left( x^2 + 4 \right) \sqrt{x^2 + 1}} \text{ dx }\]

\[\int\frac{\sin^2 x}{\cos^6 x} \text{ dx }\]

\[\int\sqrt{1 + 2x - 3 x^2}\text{  dx } \]

\[\int\frac{x^2}{\sqrt{1 - x}} \text{ dx }\]

\[\int \sin^{- 1} \sqrt{x}\ dx\]

\[\int \left( \sin^{- 1} x \right)^3 dx\]

\[\int\frac{1}{\left( x^2 + 2 \right) \left( x^2 + 5 \right)} \text{ dx}\]

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×