Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[\int\frac{\left( x + 1 \right)\left( x - 2 \right)}{\sqrt{x}}dx\]
\[ = \int \left( \frac{x^2 - 2x + x - 2}{\sqrt{x}} \right)dx\]
`=∫((x^2-x-2)/sqrt(x))dx`
\[ = \int\left( x^\frac{3}{2} - x^\frac{1}{2} - 2 x^{- \frac{1}{2}} \right)dx\]
\[ = \left[ \frac{x^\frac{3}{2} + 1}{\frac{3}{2} + 1} \right] - \left[ \frac{x^\frac{1}{2} + 1}{\frac{1}{2} + 1} \right] - 2\left[ \frac{x^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1} \right] + C\]
\[ = \frac{2}{5} x^\frac{5}{2} - \frac{2}{3} x^\frac{3}{2} - 4 x^\frac{1}{2} + C\]
APPEARS IN
संबंधित प्रश्न
\[\int\sqrt{x}\left( 3 - 5x \right) dx\]
If f' (x) = 8x3 − 2x, f(2) = 8, find f(x)
` ∫ {sec x "cosec " x}/{log ( tan x) }` dx
\[\int\sqrt{\frac{x}{1 - x}} dx\] is equal to
If \[\int\frac{1}{\left( x + 2 \right)\left( x^2 + 1 \right)}dx = a\log\left| 1 + x^2 \right| + b \tan^{- 1} x + \frac{1}{5}\log\left| x + 2 \right| + C,\] then
\[\int\frac{1}{2 - 3 \cos 2x} \text{ dx }\]
\[\int\frac{x + 3}{\left( x + 4 \right)^2} e^x dx =\]
