Advertisements
Advertisements
प्रश्न
उत्तर
\[\int\left( 2 - 3x \right) \left( 3 + 2x \right)\left( 1 - 2x \right)dx\]
\[ = \int\left( 2 - 3x \right) \left( 3 - 6x + 2x - 4 x^2 \right)dx\]
\[ = \int\left( 2 - 3x \right) \left( - 4 x^2 - 4x + 3 \right)dx\]
\[ = \int\left( - 8 x^2 - 8x + 6 + 12 x^3 + 12 x^2 - 9x \right)dx\]
\[ = \int\left( 12 x^3 + 4 x^2 - 17x + 6 \right)dx\]
\[ = \frac{12 x^4}{4} + \frac{4 x^3}{3} - \frac{17 x^2}{2} + 6x + C\]
\[ = 3 x^4 + \frac{4}{3} x^3 - \frac{17}{2} x^2 + 6x + C\]
APPEARS IN
संबंधित प्रश्न
If f' (x) = a sin x + b cos x and f' (0) = 4, f(0) = 3, f
\[\int\frac{x}{\sqrt{8 + x - x^2}} dx\]
\[\int\sqrt{\frac{x}{1 - x}} dx\] is equal to
\[\int\frac{x^3}{\sqrt{1 + x^2}}dx = a \left( 1 + x^2 \right)^\frac{3}{2} + b\sqrt{1 + x^2} + C\], then
\[\int\frac{1}{2 - 3 \cos 2x} \text{ dx }\]
\[\int {cosec}^4 2x\ dx\]
