Advertisements
Advertisements
Question
\[\int\left( x + 1 \right) \text{ log x dx }\]
Sum
Advertisements
Solution
\[\int \left( x + 1 \right)_{II} . \log_1 \text{ x dx }\]
\[ = \log x\int\left( x + 1 \right)dx - \int\left\{ \frac{d}{dx}\left( \log x \right)\int\left( x + 1 \right)dx \right\}dx\]
\[ = \log x\left[ \frac{x^2}{2} + x \right] - \int \frac{1}{x}\left( \frac{x^2}{2} + x \right)dx\]
\[ = \log x\left( \frac{x^2}{2} + x \right) - \int \left( \frac{x}{2} + 1 \right)dx\]
\[ = \log x\left( \frac{x^2}{2} + x \right) - \left( \frac{x^2}{4} + x \right) + C\]
shaalaa.com
Is there an error in this question or solution?
APPEARS IN
RELATED QUESTIONS
\[\int\frac{\left( 1 + x \right)^3}{\sqrt{x}} dx\]
\[\int \left( 3x + 4 \right)^2 dx\]
\[\int\frac{1 - \cos 2x}{1 + \cos 2x} dx\]
\[\int\frac{1}{2 - 3x} + \frac{1}{\sqrt{3x - 2}} dx\]
\[\int\frac{x^2 + 5x + 2}{x + 2} dx\]
`∫ cos ^4 2x dx `
\[\int \cos^2 \text{nx dx}\]
` ∫ cos mx cos nx dx `
` ∫ tan 2x tan 3x tan 5x dx `
\[\int\sqrt {e^x- 1} \text{dx}\]
\[\int\frac{1}{1 + x - x^2} \text{ dx }\]
\[\int\frac{e^x}{\sqrt{16 - e^{2x}}} dx\]
\[\int\frac{x^2 + x + 1}{x^2 - x + 1} \text{ dx }\]
\[\int\frac{x^3 + x^2 + 2x + 1}{x^2 - x + 1}\text{ dx }\]
\[\int\frac{\cos x}{\cos 3x} \text{ dx }\]
\[\int\frac{1}{\sin^2 x + \sin 2x} \text{ dx }\]
\[\int\frac{1}{\sqrt{3} \sin x + \cos x} dx\]
\[\int\frac{1}{1 - \cot x} dx\]
\[\int\frac{1}{1 - \tan x} \text{ dx }\]
\[\int\frac{3 + 2 \cos x + 4 \sin x}{2 \sin x + \cos x + 3} \text{ dx }\]
\[\int x \text{ sin 2x dx }\]
\[\int x\left( \frac{\sec 2x - 1}{\sec 2x + 1} \right) dx\]
\[\int x^2 \tan^{- 1} x\text{ dx }\]
\[\int\sqrt{3 - 2x - 2 x^2} \text{ dx}\]
\[\int\left( x - 2 \right) \sqrt{2 x^2 - 6x + 5} \text{ dx }\]
\[\int\frac{x^2 + 1}{x\left( x^2 - 1 \right)} dx\]
\[\int\frac{2x}{\left( x^2 + 1 \right) \left( x^2 + 3 \right)} dx\]
\[\int\frac{x^2}{\left( x^2 + 1 \right) \left( 3 x^2 + 4 \right)} dx\]
\[\int\frac{x^2 + 1}{x^4 + x^2 + 1} \text{ dx }\]
\[\int\frac{x^2 - 1}{x^4 + 1} \text{ dx }\]
\[\int\frac{x^3}{\sqrt{1 + x^2}}dx = a \left( 1 + x^2 \right)^\frac{3}{2} + b\sqrt{1 + x^2} + C\], then
\[\int x \sin^5 x^2 \cos x^2 dx\]
\[\int \sin^3 x \cos^4 x\ \text{ dx }\]
\[\int\frac{1}{a + b \tan x} \text{ dx }\]
\[\int \tan^3 x\ \sec^4 x\ dx\]
\[\int\sqrt{x^2 - a^2} \text{ dx}\]
\[\int x \sec^2 2x\ dx\]
\[\int\frac{x^5}{\sqrt{1 + x^3}} \text{ dx }\]
\[\int\frac{\sqrt{1 - \sin x}}{1 + \cos x} e^{- x/2} \text{ dx}\]
