Advertisements
Advertisements
प्रश्न
\[\int\left( 4x + 2 \right)\sqrt{x^2 + x + 1} \text{dx}\]
बेरीज
उत्तर
\[\int\left( 4x + 2 \right) \sqrt{x^2 + x + 1} \text{dx}\]
\[ = 2\int\left( 2x + 1 \right) \sqrt{x^2 + x + 1} dx\]
\[\text{Let }x^2 + x + 1 = t\]
\[ \Rightarrow \left( 2x + 1 \right) = \frac{dt}{dx}\]
\[ \Rightarrow \left( 2x + 1 \right) dx = dt\]
\[Now, 2\int\left( 2x + 1 \right) \sqrt{x^2 + x + 1} dx\]
\[ = 2\int\sqrt{t} \text{dt}\]
\[ = 2\int t^\frac{1}{2} \text{dt}\]
\[ = 2 \left[ \frac{t^\frac{1}{2} + 1}{\frac{1}{2} + 1} \right] + C\]
\[ = 2 \times \frac{2}{3} t^\frac{3}{2} + C\]
\[ = \frac{4}{3} \text{t}^\frac{3}{2} + C\]
\[ = \frac{4}{3} \left( x^2 + x + 1 \right)^\frac{3}{2} + C\]
shaalaa.com
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
APPEARS IN
संबंधित प्रश्न
` ∫ {cosec x} / {"cosec x "- cot x} ` dx
\[\int\frac{\cos x}{1 + \cos x} dx\]
\[\int\frac{x^2 + x + 5}{3x + 2} dx\]
\[\int\frac{\sec x \tan x}{3 \sec x + 5} dx\]
\[\int\frac{\cos 4x - \cos 2x}{\sin 4x - \sin 2x} dx\]
\[\int\frac{1}{\sqrt{1 - x^2}\left( 2 + 3 \sin^{- 1} x \right)} dx\]
` ∫ tan 2x tan 3x tan 5x dx `
\[\int\frac{\sin\sqrt{x}}{\sqrt{x}} dx\]
\[\int\frac{\left( \sin^{- 1} x \right)^3}{\sqrt{1 - x^2}} dx\]
\[\int\sqrt {e^x- 1} \text{dx}\]
\[\int x^2 \sqrt{x + 2} \text{ dx }\]
\[\int\frac{x^2}{\sqrt{x - 1}} dx\]
` ∫ tan^5 x sec ^4 x dx `
\[\int \cot^6 x \text{ dx }\]
` ∫ {1}/{a^2 x^2- b^2}dx`
\[\int\frac{1}{\sqrt{\left( 2 - x \right)^2 + 1}} dx\]
\[\int\frac{e^x}{\left( 1 + e^x \right)\left( 2 + e^x \right)} dx\]
\[\int\frac{1}{\sqrt{5 - 4x - 2 x^2}} dx\]
\[\int\frac{\sin 2x}{\sqrt{\cos^4 x - \sin^2 x + 2}} dx\]
\[\int\frac{x - 1}{3 x^2 - 4x + 3} dx\]
\[\int\frac{x^2 + x + 1}{x^2 - x + 1} \text{ dx }\]
\[\int\frac{x - 1}{\sqrt{x^2 + 1}} \text{ dx }\]
\[\int e^x \left( \cot x - {cosec}^2 x \right) dx\]
\[\int\sqrt{3 - x^2} \text{ dx}\]
\[\int\left( x + 1 \right) \sqrt{x^2 + x + 1} \text{ dx }\]
\[\int\frac{1}{x\left( x - 2 \right) \left( x - 4 \right)} dx\]
\[\int\frac{1}{\left( x - 1 \right) \left( x + 1 \right) \left( x + 2 \right)} dx\]
\[\int\frac{5}{\left( x^2 + 1 \right) \left( x + 2 \right)} dx\]
\[\int\frac{\left( x^2 + 1 \right) \left( x^2 + 2 \right)}{\left( x^2 + 3 \right) \left( x^2 + 4 \right)} dx\]
\[\int\frac{1}{\left( 1 + x^2 \right) \sqrt{1 - x^2}} \text{ dx }\]
\[\int\frac{x}{\left( x^2 + 4 \right) \sqrt{x^2 + 9}} \text{ dx}\]
\[\int\frac{1}{\cos x + \sqrt{3} \sin x} \text{ dx } \] is equal to
` \int \text{ x} \text{ sec x}^2 \text{ dx is equal to }`
\[\int\frac{x^2}{\left( x - 1 \right)^3} dx\]
\[\int\frac{1 + \sin x}{\sin x \left( 1 + \cos x \right)} \text{ dx }\]
\[\int\frac{\sin^2 x}{\cos^6 x} \text{ dx }\]
\[\int e^x \frac{\left( 1 - x \right)^2}{\left( 1 + x^2 \right)^2} \text{ dx }\]
\[\int\frac{\sqrt{1 - \sin x}}{1 + \cos x} e^{- x/2} \text{ dx}\]
