Advertisements
Advertisements
प्रश्न
\[\int 5^{x + \tan^{- 1} x} . \left( \frac{x^2 + 2}{x^2 + 1} \right) dx\]
योग
उत्तर
\[\int 5^{x + \tan^{- 1} x} \cdot \left( \frac{x^2 + 2}{x^2 + 1} \right)dx\]
\[\text{Let x} + \tan^{- 1} x = t\]
\[\left( 1 + \frac{1}{1 + x^2} \right) = \frac{dt}{dx}\]
\[ \Rightarrow \left( \frac{x^2 + 1 + 1}{x^2 + 1} \right)dx = dt\]
\[ \Rightarrow \left( \frac{x^2 + 2}{x^2 + 1} \right)dx = dt\]
\[Now, \int 5^{x + \tan^{- 1} x} \cdot \left( \frac{x^2 + 2}{x^2 + 1} \right)dx\]
\[ = \int 5^t dt\]
\[ = \frac{5^t}{\log 5} + C\]
\[ = \frac{5^{x + \tan^1 x}}{\log 5} + C\]
shaalaa.com
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
APPEARS IN
संबंधित प्रश्न
\[\int\left( 3x\sqrt{x} + 4\sqrt{x} + 5 \right)dx\]
\[\int \left( 3x + 4 \right)^2 dx\]
\[\int\frac{x + 1}{\sqrt{2x + 3}} dx\]
\[\int\frac{e^x + 1}{e^x + x} dx\]
\[\int\frac{x + 1}{x \left( x + \log x \right)} dx\]
\[\int \sin^4 x \cos^3 x \text{ dx }\]
\[\int\frac{x^2 - 1}{x^2 + 4} dx\]
\[\int\frac{\sec^2 x}{1 - \tan^2 x} dx\]
\[\int\frac{x}{\sqrt{4 - x^4}} dx\]
\[\int\frac{x^2 + x - 1}{x^2 + x - 6}\text{ dx }\]
\[\int\frac{\left( 1 - x^2 \right)}{x \left( 1 - 2x \right)} \text
{dx\]
\[\int\frac{6x - 5}{\sqrt{3 x^2 - 5x + 1}} \text{ dx }\]
\[\int\frac{x}{\sqrt{x^2 + x + 1}} \text{ dx }\]
\[\int\frac{1}{3 + 2 \cos^2 x} \text{ dx }\]
\[\int\frac{1}{\sqrt{3} \sin x + \cos x} dx\]
\[\int\frac{1}{1 - \tan x} \text{ dx }\]
\[\int \tan^{- 1} \left( \frac{2x}{1 - x^2} \right) \text{ dx }\]
\[\int e^x \left( \cot x + \log \sin x \right) dx\]
\[\int e^x \frac{\left( 1 - x \right)^2}{\left( 1 + x^2 \right)^2} \text{ dx }\]
\[\int e^x \left( \log x + \frac{1}{x} \right) dx\]
\[\int e^x \cdot \frac{\sqrt{1 - x^2} \sin^{- 1} x + 1}{\sqrt{1 - x^2}} \text{ dx }\]
\[\int\left( 2x - 5 \right) \sqrt{2 + 3x - x^2} \text{ dx }\]
\[\int\frac{3 + 4x - x^2}{\left( x + 2 \right) \left( x - 1 \right)} dx\]
\[\int\frac{x^2 + 1}{x\left( x^2 - 1 \right)} dx\]
\[\int\frac{5 x^2 - 1}{x \left( x - 1 \right) \left( x + 1 \right)} dx\]
\[\int\frac{1}{\cos x \left( 5 - 4 \sin x \right)} dx\]
\[\int\frac{1}{\left( x - 1 \right) \sqrt{x + 2}} \text{ dx }\]
\[\int\frac{1}{\left( x + 1 \right) \sqrt{x^2 + x + 1}} \text{ dx }\]
` \int \text{ x} \text{ sec x}^2 \text{ dx is equal to }`
\[\int x^{\sin x} \left( \frac{\sin x}{x} + \cos x . \log x \right) dx\] is equal to
\[\int\frac{1}{\sqrt{x} + \sqrt{x + 1}} \text{ dx }\]
\[\int\frac{\sin 2x}{a^2 + b^2 \sin^2 x}\]
\[\int \cos^5 x\ dx\]
\[\int\frac{\sin^5 x}{\cos^4 x} \text{ dx }\]
\[\int\sqrt{a^2 + x^2} \text{ dx }\]
\[\int \left( \sin^{- 1} x \right)^3 dx\]
\[\int\frac{x^2}{\left( x - 1 \right)^3 \left( x + 1 \right)} \text{ dx}\]
Evaluate : \[\int\frac{\cos 2x + 2 \sin^2 x}{\cos^2 x}dx\] .
