मराठी

∫ X 2 + X − 1 X 2 + X − 6 D X - Mathematics

Advertisements
Advertisements

प्रश्न

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

उत्तर

\[\int\left( \frac{x^2 + x - 1}{x^2 + x - 6} \right)dx\]
\[\frac{x^2 + x - 1}{x^2 + x - 6} = 1 + \frac{5}{x^2 + x - 6}\]
\[ \int\left( \frac{x^2 + x - 1}{x^2 + x - 6} \right)dx\]
\[ = ∫ dx + 5\int\frac{dx}{x^2 + x - 6}\]
\[ = ∫  dx + 5\int\frac{dx}{x^2 + x + \left( \frac{1}{2} \right)^2 - \left( \frac{1}{2} \right)^2 - 6}\]
\[ = ∫ dx + 5\int\frac{dx}{\left( x + \frac{1}{2} \right)^2 - \frac{1}{4} - 6}\]


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


\[ = x + 5 \times \frac{1}{2 \times \frac{5}{2}} \text{ log } \left| \frac{x + \frac{1}{2} - \frac{5}{2}}{x + \frac{1}{2} + \frac{5}{2}} \right| + C\]
\[ = x + \text{ log } \left| \frac{x - 2}{x + 3} \right| + C\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 19: Indefinite Integrals - Exercise 19.2 [पृष्ठ १०६]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 19 Indefinite Integrals
Exercise 19.2 | Q 2 | पृष्ठ १०६

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

If f' (x) = x − \[\frac{1}{x^2}\]  and  f (1)  \[\frac{1}{2},    find  f(x)\]

 


\[\int \tan^{3/2} x \sec^2 \text{x dx}\]

\[\int\frac{\sin\sqrt{x}}{\sqrt{x}} dx\]

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

\[\int\frac{1}{x^2 \left( x^4 + 1 \right)^{3/4}} dx\]

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

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

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

\[\int\frac{1}{\sqrt{16 - 6x - x^2}} dx\]

\[\int\frac{\sin 8x}{\sqrt{9 + \sin^4 4x}} dx\]

\[\int\frac{\cos x - \sin x}{\sqrt{8 - \sin2x}}dx\]

\[\int\frac{1 - 3x}{3 x^2 + 4x + 2}\text{  dx}\]

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

\[\int\frac{x^2}{x^2 + 6x + 12} \text{ dx }\]

\[\int\frac{2}{2 + \sin 2x}\text{ dx }\]

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

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

\[\int\frac{1}{\sin x + \sqrt{3} \cos x} \text{ dx  }\]

\[\int\frac{1}{\sqrt{3} \sin x + \cos x} dx\]

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

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

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

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

\[\int e^x \left( \tan x - \log \cos x \right) dx\]

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

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

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

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

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

The primitive of the function \[f\left( x \right) = \left( 1 - \frac{1}{x^2} \right) a^{x + \frac{1}{x}} , a > 0\text{ is}\]


\[\int\sqrt{\frac{x}{1 - x}} dx\]  is equal to


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


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

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

\[\int\frac{1}{5 - 4 \sin x} \text{ dx }\]

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

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

Find : \[\int\frac{dx}{\sqrt{3 - 2x - x^2}}\] .


Find :  \[\int\frac{e^x}{\left( 2 + e^x \right)\left( 4 + e^{2x} \right)}dx.\] 

 


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×