मराठी

∫ Tan − 1 ( 3 X − X 3 1 − 3 X 2 ) D X - Mathematics

Advertisements
Advertisements

प्रश्न

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

उत्तर

\[\text{ Let I } = \int \tan^{- 1} \left( \frac{3x - x^3}{1 - 3 x^2} \right) \text{ dx }\]
\[ = \int3 \tan^{- 1} \left( x \right) \text{ dx }\]
\[ = 3\int\left[ \tan^{- 1} \left( x \right) \times 1 \right] \text{ dx }\]
\[ = 3 \left[ \tan^{- 1} x \times x - \int\frac{1}{1 + x^2} \times\text{  x dx } \right]\]
\[ = 3x \tan^{- 1} x - 3\int\frac{x}{1 + x^2} dx\]
\[\text{ let 1 }+ x^2 = t\]
\[ \Rightarrow \text{ 2x dx }= dt\]
\[\text{ Then,} \]
\[I = 3x \tan^{- 1} x - \frac{3}{2}\int\frac{dt}{t}\]
\[ = 3x \tan^{- 1} x - \frac{3}{2} \text{ log } \left| t \right| + C\]
\[ = 3x \tan^{- 1} x - \frac{3}{2} \text{ log} \left| 1 + x^2 \right| + C\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 19: Indefinite Integrals - Exercise 19.25 [पृष्ठ १३४]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 19 Indefinite Integrals
Exercise 19.25 | Q 37 | पृष्ठ १३४

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

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

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

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

\[\int\frac{x^3 - 3 x^2 + 5x - 7 + x^2 a^x}{2 x^2} dx\]

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)
 

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

\[\int \tan^2 \left( 2x - 3 \right) dx\]


` ∫   sin x  \sqrt (1-cos 2x)    dx `

 


` ∫    cos  mx  cos  nx  dx `

 


\[\int\frac{\sec x \tan x}{3 \sec x + 5} dx\]

\[\int\frac{1}{x (3 + \log x)} dx\]

\[\int\frac{\tan x}{\sqrt{\cos x}} dx\]

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

 


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

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

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

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

\[\int\frac{1}{4 x^2 + 12x + 5} dx\]

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

\[\int\frac{1}{\sqrt{\left( 1 - x^2 \right)\left\{ 9 + \left( \sin^{- 1} x \right)^2 \right\}}} dx\]

` ∫  {x-3} /{ x^2 + 2x - 4 } dx `


\[\int\frac{a x^3 + bx}{x^4 + c^2} dx\]

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

`int"x"^"n"."log"  "x"  "dx"`

\[\int x^3 \tan^{- 1}\text{  x dx }\]

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

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

Evaluate the following integral:

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

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

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

\[\int \cos^3 (3x)\ dx\]

\[\int \tan^4 x\ dx\]

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

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

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

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

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

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

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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×