Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[\int\left( \frac{2 x^4 + 7 x^3 + 6 x^2}{x^2 + 2x} \right)dx\]
\[ = \int \frac{x^2 \left( 2 x^2 + 7x + 6 \right)}{x\left( x + 2 \right)}dx\]
\[ = \int\frac{x\left[ 2 x^2 + 4x + 3x + 6 \right]}{x + 2}dx\]
\[ = \int\frac{x\left( 2x\left( x + 2 \right) + 3\left( x + 2 \right) \right)}{\left( x + 2 \right)}dx\]
`= c x ( (2x+3)(x+2)) / (x+2) dx`
\[ = \int\left( 2 x^2 + 3x \right)dx\]
` = 2 ∫ x^2 dx + 3∫ x dx`
\[ = 2\left[ \frac{x^3}{3} \right] + 3\left[ \frac{x^2}{2} \right] + C\]
\[ = \frac{2}{3} x^3 + \frac{3}{2} x^2 + C\]
APPEARS IN
संबंधित प्रश्न
\[\int\sqrt{x}\left( 3 - 5x \right) dx\]
If f' (x) = x − \[\frac{1}{x^2}\] and f (1) \[\frac{1}{2}, find f(x)\]
If \[\int\frac{\sin^8 x - \cos^8 x}{1 - 2 \sin^2 x \cos^2 x} dx\]
\[\int\sqrt{\frac{x}{1 - x}} dx\] is equal to
\[\int\frac{1 - x^4}{1 - x} \text{ dx }\]
\[\int\frac{1}{2 + \cos x} \text{ dx }\]
\[\int\frac{x + 3}{\left( x + 4 \right)^2} e^x dx =\]
