Practicing Success
Practicing Success
CUET Preparation Today
If $\int \frac{d x}{\sqrt{x+2}-\sqrt{x+1}}=\frac{2}{3}\left[(\lambda+1)^{\frac{3}{2}}+\lambda^{\frac{3}{2}}\right]+C$, then the value of λ is |
x - 1 x x + 1 \(\frac{1}{x}\) |
x + 1 |
$I=\int \frac{d x}{\sqrt{x+2}-\sqrt{x+1}}=\int(\sqrt{x+2}+\sqrt{x+1})dx$ $=\frac{2}{3}(x+2)^{\frac{3}{2}}+\frac{2}{3}(x+1)^{\frac{3}{2}}+C$ = x + 1 |
