मराठी

∫ 1 √ X + 4 √ X D X - Mathematics

Advertisements
Advertisements

प्रश्न

\[\int\frac{1}{\sqrt{x} + \sqrt[4]{x}}dx\]
बेरीज

उत्तर

\[\text{ Let  I } = \int\frac{1}{\sqrt{x} + \sqrt[4]{x}}dx\]
\[ \text{ Let  x }= t^4 \]
\[ \text{On differentiating both sides, we get}\]
\[ dx = 4 t^3 dt\]
\[ \therefore I = \int\frac{4 t^3}{\sqrt{t^4} + \sqrt[4]{t^4}}dt\]
\[ = \int\frac{4 t^3}{t^2 + t}dt\]
\[ = 4\int\frac{t^2}{t + 1}dt\]


\[ = 4\int\frac{\left( t - 1 \right)\left( t + 1 \right) + 1}{t + 1}dt\]
\[ = 4\int\left[ \left( t - 1 \right) + \frac{1}{t + 1} \right]dt\]
\[ = 4\left[ \frac{t^2}{2} - t + \log\left( t + 1 \right) \right] + c\]
\[ = 2\sqrt{x} - 4\sqrt[4]{x} + 4 \log\left( \sqrt[4]{x} + 1 \right) + c\]
\[Hence, \int\frac{1}{\sqrt{x} + \sqrt[4]{x}}dx = 2\sqrt{x} - 4\sqrt[4]{x} + 4 \log\left( \sqrt[4]{x} + 1 \right) + c\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 19: Indefinite Integrals - Exercise 19.10 [पृष्ठ ६५]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 19 Indefinite Integrals
Exercise 19.10 | Q 9 | पृष्ठ ६५

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

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

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

If f' (x) = a sin x + b cos x and f' (0) = 4, f(0) = 3, f

\[\left( \frac{\pi}{2} \right)\] = 5, find f(x)
 

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

` ∫   cos  3x   cos  4x` dx  

\[\int\text{sin mx }\text{cos nx dx m }\neq n\]

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

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

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

\[\int \sin^7 x  \text{ dx }\]

Evaluate the following integrals:
\[\int\frac{x^2}{\left( a^2 - x^2 \right)^{3/2}}dx\]

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

\[\int\frac{\cos x}{\sqrt{\sin^2 x - 2 \sin x - 3}} dx\]

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

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

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

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

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

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


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

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

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

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

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

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

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

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

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

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

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

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

\[\int\frac{1}{7 + 5 \cos x} dx =\]

If \[\int\frac{1}{\left( x + 2 \right)\left( x^2 + 1 \right)}dx = a\log\left| 1 + x^2 \right| + b \tan^{- 1} x + \frac{1}{5}\log\left| x + 2 \right| + C,\] then


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

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

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

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

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

\[\int {cosec}^4 2x\ dx\]


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×