मराठी

∫ ( X + 1 ) √ X 2 − X + 1 D X - Mathematics

Advertisements
Advertisements

प्रश्न

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

उत्तर

\[\text{ Let I } = \int \left( x + 1 \right) \sqrt{x^2 - x + 1} \text{ dx}\]
\[\text{ Also }, x + 1 = \lambda\frac{d}{dx}\left( x^2 - x + 1 \right) + \mu\]
\[ \Rightarrow x + 1 = \lambda\left( 2x - 1 \right) + \mu\]
\[ \Rightarrow x + 1 = \lambda\left( 2x - 1 \right) + \mu\]
\[ \Rightarrow x + 1 = \left( 2\lambda \right)x + \mu - \lambda\]
\[\text{Equating the coefficient of like terms}\]
\[2\lambda = 1\]
\[ \Rightarrow \lambda = \frac{1}{2}\]
\[\text{ And }\]
\[\mu - \lambda = 1\]
\[ \Rightarrow \mu - \frac{1}{2} = 1\]
\[ \Rightarrow \mu = \frac{3}{2}\]
\[ \therefore I = \int\left[ \frac{1}{2}\left( 2x - 1 \right) + \frac{3}{2} \right] \sqrt{x^2 - x + 1}dx\]
\[ = \frac{1}{2}\int\left( 2x - 1 \right) \sqrt{x^2 - x + 1}dx + \frac{3}{2}\int\sqrt{x^2 - x + 1}dx\]
\[ = \frac{1}{2}\int\left( 2x - 1 \right) \sqrt{x^2 - x + 1} \text{ dx}+ \frac{3}{2}\int \sqrt{x^2 - x + \frac{1}{4} - \frac{1}{4} + 1} \text{ dx}\]
\[ = \frac{1}{2}\int\left( 2x - 1 \right) \sqrt{x^2 - x + 1} \text{ dx} + \frac{3}{2}\int\sqrt{\left( x - \frac{1}{2} \right)^2 + \frac{3}{4}}\text{ dx}\]
\[ = \frac{1}{2}\int\left( 2x - 1 \right) \sqrt{x^2 - x + 1} \text{ dx}+ \frac{3}{2}\int\sqrt{\left( x - \frac{1}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2}\text{ dx}\]
\[\text{ Let x}^2 - x + 1 = t\]
\[ \Rightarrow \left( 2x - 1 \right)dx = dt\]
\[ \therefore I = \frac{1}{2}\int\sqrt{t} \text{ dt }+ \frac{3}{2}\int\sqrt{\left( x - \frac{1}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2} dx\]
\[ = \frac{1}{2} \times \left( \frac{t^\frac{3}{2}}{\frac{3}{2}} \right) + \frac{3}{2}\left[ \frac{x - \frac{1}{2}}{2} \sqrt{\left( x - \frac{1}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2} + \frac{3}{8}\text{ log } \left| x - \frac{1}{2} + \sqrt{\left( x - \frac{1}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2} \right| \right] + C\]
\[ = \frac{1}{3} \left( x^2 - x + 1 \right)^\frac{3}{2} + \frac{3}{8}\left( 2x - 1 \right) \sqrt{x^2 - x + 1} + \frac{9}{16}\text{ log } \left| x - \frac{1}{2} + \sqrt{x^2 - x + 1} \right| + C\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 19: Indefinite Integrals - Exercise 19.29 [पृष्ठ १५८]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 19 Indefinite Integrals
Exercise 19.29 | Q 1 | पृष्ठ १५८

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

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

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

` ∫   e^{m   sin ^-1  x}/ \sqrt{1-x^2}  ` dx

 


\[\int\frac{\sec^2 \sqrt{x}}{\sqrt{x}} dx\]

\[\int 5^{5^{5^x}} 5^{5^x} 5^x dx\]

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

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

\[\int x \cos^3 x^2 \sin x^2 \text{ dx }\]

\[\int\frac{e^x}{e^{2x} + 5 e^x + 6} dx\]

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

\[\int\frac{1}{\sqrt{7 - 6x - x^2}} dx\]

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

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

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

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

`int 1/(cos x - sin x)dx`

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

\[\int x^3 \cos x^2 dx\]

\[\int \log_{10} x\ dx\]

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

\[\int \cos^{- 1} \left( 4 x^3 - 3x \right) \text{ dx }\]

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

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

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

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

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

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

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


\[\int\frac{\cos2x - \cos2\theta}{\cos x - \cos\theta}dx\] is equal to 

\[\int\sin x \sin 2x \text{ sin  3x  dx }\]


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

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

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

\[\int\frac{1}{3 x^2 + 13x - 10} \text{ dx }\]

\[\int\frac{5x + 7}{\sqrt{\left( x - 5 \right) \left( x - 4 \right)}} \text{ dx }\]

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

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

\[\int\frac{x^5}{\sqrt{1 + x^3}} \text{ dx }\]

\[\int x^2 \tan^{- 1} x\ dx\]

Find: `int (3x +5)/(x^2+3x-18)dx.`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×