Advertisements
Advertisements
प्रश्न
\[\int\left( x + 1 \right) \text{ log x dx }\]
बेरीज
Advertisements
उत्तर
\[\int \left( x + 1 \right)_{II} . \log_1 \text{ x dx }\]
\[ = \log x\int\left( x + 1 \right)dx - \int\left\{ \frac{d}{dx}\left( \log x \right)\int\left( x + 1 \right)dx \right\}dx\]
\[ = \log x\left[ \frac{x^2}{2} + x \right] - \int \frac{1}{x}\left( \frac{x^2}{2} + x \right)dx\]
\[ = \log x\left( \frac{x^2}{2} + x \right) - \int \left( \frac{x}{2} + 1 \right)dx\]
\[ = \log x\left( \frac{x^2}{2} + x \right) - \left( \frac{x^2}{4} + x \right) + C\]
shaalaa.com
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
APPEARS IN
संबंधित प्रश्न
\[\int\frac{x^5 + x^{- 2} + 2}{x^2} dx\]
\[\int \left( 3x + 4 \right)^2 dx\]
\[\int \sin^{- 1} \left( \frac{2 \tan x}{1 + \tan^2 x} \right) dx\]
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)
` ∫ 1/ {1+ cos 3x} ` dx
\[\int\sqrt{\frac{1 + \cos 2x}{1 - \cos 2x}} dx\]
\[\int\frac{\text{sin} \left( x - \alpha \right)}{\text{sin }\left( x + \alpha \right)} dx\]
\[\int\frac{e^x + 1}{e^x + x} dx\]
\[\int\frac{2 \cos 2x + \sec^2 x}{\sin 2x + \tan x - 5} dx\]
\[\int \tan^3 \text{2x sec 2x dx}\]
\[\int 5^{5^{5^x}} 5^{5^x} 5^x dx\]
\[\int\frac{x^2}{\sqrt{1 - x}} dx\]
\[\int\frac{1}{\sqrt{x} + \sqrt[4]{x}}dx\]
\[\int {cosec}^4 \text{ 3x } \text{ dx } \]
\[\int\frac{e^{3x}}{4 e^{6x} - 9} dx\]
\[\int\frac{e^x}{\sqrt{16 - e^{2x}}} dx\]
\[\int\frac{\cos x}{\sqrt{4 - \sin^2 x}} dx\]
` ∫ \sqrt{"cosec x"- 1} dx `
\[\int\frac{x - 1}{\sqrt{x^2 + 1}} \text{ dx }\]
\[\int\frac{1}{3 + 2 \cos^2 x} \text{ dx }\]
\[\int\frac{1}{1 - \tan x} \text{ dx }\]
\[\int e^x \left( \cot x - {cosec}^2 x \right) dx\]
\[\int e^x \frac{x - 1}{\left( x + 1 \right)^3} \text{ dx }\]
\[\int\left( 2x + 3 \right) \sqrt{x^2 + 4x + 3} \text{ dx }\]
\[\int\frac{2x + 1}{\left( x - 2 \right) \left( x - 3 \right)} dx\]
\[\int\frac{x^2 + 1}{x^4 + x^2 + 1} \text{ dx }\]
\[\int\frac{1}{x^4 + x^2 + 1} \text{ dx }\]
\[\int\frac{1}{\left( x^2 + 1 \right) \sqrt{x}} \text{ dx }\]
\[\int\frac{\cos 2x - 1}{\cos 2x + 1} dx =\]
\[\int\frac{1}{e^x + 1} \text{ dx }\]
\[\int\sqrt{\sin x} \cos^3 x\ \text{ dx }\]
\[\int\frac{1}{a + b \tan x} \text{ dx }\]
\[\int\frac{1}{1 + 2 \cos x} \text{ dx }\]
\[\int\frac{1 + \sin x}{\sin x \left( 1 + \cos x \right)} \text{ dx }\]
\[\int\frac{1}{\sec x + cosec x}\text{ dx }\]
\[\int\frac{\sin x + \cos x}{\sin^4 x + \cos^4 x} \text{ dx }\]
\[\int \sin^{- 1} \sqrt{x}\ dx\]
\[\int\frac{x}{x^3 - 1} \text{ dx}\]
