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

\[\int\left( \frac{m}{x} + \frac{x}{m} + m^x + x^m + mx \right) dx\]

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

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

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

\[\int\frac{1}{\sqrt{x + a} + \sqrt{x + b}} dx\]

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

` ∫  1/ {1+ cos   3x}  ` dx


` ∫    cos  mx  cos  nx  dx `

 


\[\int\frac{a}{b + c e^x} dx\]

\[\int\frac{\sin 2x}{\sin \left( x - \frac{\pi}{6} \right) \sin \left( x + \frac{\pi}{6} \right)} dx\]

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

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

\[\int\frac{\sin^5 x}{\cos^4 x} \text{ dx }\]

\[\int \cos^7 x \text{ dx  } \]

Evaluate the following integrals:

\[\int\cos\left\{ 2 \cot^{- 1} \sqrt{\frac{1 + x}{1 - x}} \right\}dx\]

\[\int\frac{1}{a^2 - b^2 x^2} dx\]

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

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

\[\int\frac{1}{\sqrt{8 + 3x - x^2}} dx\]

`  ∫ \sqrt{"cosec x"- 1}  dx `

\[\int\frac{\left( 3\sin x - 2 \right)\cos x}{13 - \cos^2 x - 7\sin x}dx\]

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

\[\int\frac{1}{1 - \cot x} dx\]

\[\int e^x \left( \frac{\sin 4x - 4}{1 - \cos 4x} \right) dx\]

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

\[\int\frac{\sqrt{16 + \left( \log x \right)^2}}{x} \text{ dx}\]

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

\[\int x\sqrt{x^2 + x} \text{  dx }\]

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

\[\int\frac{x^2 + 9}{x^4 + 81} \text{ dx }\]

 


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

If \[\int\frac{\sin^8 x - \cos^8 x}{1 - 2 \sin^2 x \cos^2 x} dx\]


\[\int\frac{x^3}{x + 1}dx\] is equal to

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

\[\int \tan^3 x\ dx\]

\[\int \tan^5 x\ dx\]

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

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

Evaluate : \[\int\frac{\cos 2x + 2 \sin^2 x}{\cos^2 x}dx\] .


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×