मराठी

∫ X + 3 ( X + 1 ) 4 D X - Mathematics

Advertisements
Advertisements

प्रश्न

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

उत्तर

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

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 19: Indefinite Integrals - Exercise 19.03 [पृष्ठ २३]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 19 Indefinite Integrals
Exercise 19.03 | Q 4 | पृष्ठ २३

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

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


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

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

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

\[\int\frac{\tan x}{\sec x + \tan x} dx\]

` ∫  {cosec x} / {"cosec x "- cot x} ` dx      


If f' (x) = x + bf(1) = 5, f(2) = 13, find f(x)


\[\int \left( 2x - 3 \right)^5 + \sqrt{3x + 2}  \text{dx} \]

\[\int\left( 5x + 3 \right) \sqrt{2x - 1} dx\]

` ∫   cos  3x   cos  4x` dx  

\[\int\frac{\sin 2x}{a^2 + b^2 \sin^2 x} dx\]

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

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

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

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

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

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

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

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

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

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

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

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

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

\[\int\left( 2x + 3 \right) \sqrt{x^2 + 4x + 3} \text{  dx }\]

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

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

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

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

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

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

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

The value of \[\int\frac{\sin x + \cos x}{\sqrt{1 - \sin 2x}} dx\] is equal to


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


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

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

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

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

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

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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×