Advertisements
Advertisements
प्रश्न
उत्तर
\[\int\frac{x^2}{\sqrt{x - 1}}\text{ dx }\]
\[\text{Let x} - 1 = t^2 \]
\[ \Rightarrow x = t^2 + 1\]
\[ \Rightarrow 1 = 2t \frac{dt}{dx}\]
\[ \Rightarrow dx = \text{ 2t dt }\]
\[Now, \int\frac{x^2}{\sqrt{x - 1}}\text{ dx }\]
\[ = \int\frac{\left( t^2 + 1 \right)^2}{t}\text{ 2t dt }\]
\[ = 2\int\left( t^4 + 2 t^2 + 1 \right)dt\]
\[ = 2\left[ \frac{t^{4 + 1}}{4 + 1} + \frac{2 t^{2 + 1}}{2 + 1} + t \right] + C\]
\[ = 2\left[ \frac{t^5}{5} + \frac{2 t^3}{3} + t \right] + C\]
\[ = 2\left[ \frac{3 t^5 + 10 t^3 + 15t}{15} \right] + C\]
\[ = \frac{2}{15}t\left[ 3 t^4 + 10 t^2 + 15 \right] + C\]
\[ = \frac{2}{15}\sqrt{x - 1} \left[ 3 \left( x - 1 \right)^2 + 10\left( x - 1 \right) + 15 \right] + C\]
\[ = \frac{2}{15}\sqrt{x - 1} \left[ 3\left( x^2 - 2x + 1 \right) + 10x - 10 + 15 \right] + C\]
\[ = \frac{2}{15}\sqrt{x - 1} \left[ 3 x^2 - 6x + 3 + 10x - 10 + 15 \right] + C\]
\[ = \frac{2}{15}\sqrt{x - 1}\left[ 3 x^2 + 4x + 8 \right] + C\]
APPEARS IN
संबंधित प्रश्न
If f' (x) = 8x3 − 2x, f(2) = 8, find f(x)
If f' (x) = a sin x + b cos x and f' (0) = 4, f(0) = 3, f
`∫ cos ^4 2x dx `
` ∫ e^{m sin ^-1 x}/ \sqrt{1-x^2} ` 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
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
\[\int\frac{x + 2}{\left( x + 1 \right)^3} \text{ dx }\]
\[\int\sqrt{\frac{1 - x}{x}} \text{ dx}\]
