Advertisements
Advertisements
प्रश्न
\[\int\frac{e^x + 1}{e^x + x} dx\]
बेरीज
उत्तर
\[\text{Let I} = \int\frac{e^x + 1}{e^x + x}dx\]
\[\text{Putting }e^x + x = t\]
\[ \Rightarrow e^x + 1 = \frac{dt}{dx}\]
\[ \Rightarrow \left( e^x + 1 \right)dx = dt\]
\[ \therefore I = \int\frac{1}{t}dt\]
\[ = \text{ln }\left| t \right| + C\]
\[ = \text{ln }\left| e^x + x \right| + C \left[ \because t = e^x + x \right]\]
shaalaa.com
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
APPEARS IN
संबंधित प्रश्न
\[\int\frac{1 - \cos x}{1 + \cos x} dx\]
\[\int\frac{1}{\sqrt{x + 1} + \sqrt{x}} dx\]
` ∫ cos mx cos nx dx `
\[\int\frac{\text{sin} \left( x - a \right)}{\text{sin}\left( x - b \right)} dx\]
\[\int\frac{\sin 2x}{\left( a + b \cos 2x \right)^2} dx\]
\[\ \int\ x \left( 1 - x \right)^{23} dx\]
` ∫ tan^5 x dx `
\[\int\frac{1}{a^2 x^2 + b^2} dx\]
\[\int\frac{1}{\sqrt{a^2 + b^2 x^2}} dx\]
\[\int\frac{1}{\sqrt{16 - 6x - x^2}} dx\]
\[\int\frac{x^2 + x + 1}{x^2 - x} dx\]
\[\int\frac{\sin 2x}{\sin^4 x + \cos^4 x} \text{ dx }\]
`int 1/(cos x - sin x)dx`
\[\int\frac{1}{\sqrt{3} \sin x + \cos x} dx\]
\[\int\frac{1}{1 - \cot x} dx\]
\[\int\frac{x + \sin x}{1 + \cos x} \text{ dx }\]
\[\int \sec^{- 1} \sqrt{x}\ dx\]
\[\int e^x \left( \cos x - \sin x \right) dx\]
\[\int e^x \frac{x - 1}{\left( x + 1 \right)^3} \text{ dx }\]
\[\int\frac{e^x \left( x - 4 \right)}{\left( x - 2 \right)^3} \text{ dx }\]
\[\int\frac{\sin 2x}{\left( 1 + \sin x \right) \left( 2 + \sin x \right)} dx\]
\[\int\frac{x}{\left( x^2 - a^2 \right) \left( x^2 - b^2 \right)} dx\]
\[\int\frac{x^2 + 1}{\left( x - 2 \right)^2 \left( x + 3 \right)} dx\]
\[\int\frac{2x + 1}{\left( x - 2 \right) \left( x - 3 \right)} dx\]
\[\int\frac{1}{\left( x - 1 \right) \sqrt{2x + 3}} \text{ dx }\]
\[\int\frac{x}{\left( x - 3 \right) \sqrt{x + 1}} dx\]
The primitive of the function \[f\left( x \right) = \left( 1 - \frac{1}{x^2} \right) a^{x + \frac{1}{x}} , a > 0\text{ is}\]
\[\int\frac{x^9}{\left( 4 x^2 + 1 \right)^6}dx\] is equal to
\[\int \cos^3 (3x)\ dx\]
\[\int \tan^4 x\ dx\]
\[\int\frac{1}{\sqrt{x^2 - a^2}} \text{ dx }\]
\[\int\frac{1}{4 x^2 + 4x + 5} dx\]
\[\int\frac{\sin^6 x}{\cos x} \text{ dx }\]
\[\int \tan^5 x\ \sec^3 x\ dx\]
\[\int\frac{1}{x\sqrt{1 + x^3}} \text{ dx}\]
\[\int\frac{\sqrt{1 - \sin x}}{1 + \cos x} e^{- x/2} \text{ dx}\]
\[\int\frac{\sin 4x - 2}{1 - \cos 4x} e^{2x} \text{ dx}\]
\[\int\frac{x^2 + 1}{x^2 - 5x + 6} \text{ dx }\]
\[\int\frac{x^2}{x^2 + 7x + 10}\text{ dx }\]
