$\int_1^2 \frac{\mathrm{d} x}{\left(x^2-2 x+4\right)^{\frac{3}{2}}}=\frac{\mathrm{k}}{\mathrm{k}+5}$, then $\mathrm{k}$ has the value

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Aicharya · Accepted Answer

$	ext { Let } \begin{aligned} \mathrm{I} & =\int_1^2 \frac{\mathrm{d} x}{\left(x^2-2 x+4ight)^{\frac{3}{2}}} \ & =\int_1^2 \frac{\mathrm{d} x}{\left[(x-1)^2+3ight]^{\frac{3}{2}}} \end{aligned}$ Put $x-1=\sqrt{3} 	an 	heta$ $\mathrm{d} x=\sqrt{3} \sec ^2 	heta \mathrm{d} 	heta$ When $x=1, 	heta=0$ When $x=2, 	heta=\frac{\pi}{6}$ $\begin{aligned} & 	herefore \quad I=\int_0^{\frac{\pi}{6}} \frac{\sqrt{3} \sec ^2 	heta}{\left[3 	an ^2 	heta+3ight]^{\frac{3}{2}}} d 	heta \ & =\int_0^{\frac{\pi}{6}} \frac{\sqrt{3} \sec ^2 	heta}{\left[3\left(1+	an ^2 	hetaight)ight]^{\frac{3}{2}}} \ & =\int_0^{\frac{\pi}{6}} \frac{\sqrt{3} \sec ^2 	heta}{3 \cdot \sqrt{3}\left(\sec ^2 	hetaight)^{\frac{3}{2}}} \ & =\int_0^{\frac{\pi}{6}} \frac{1}{3} \cdot \frac{\sec ^2 	heta}{\sec ^3 	heta} \ & =\frac{1}{3} \int_{\infty}^{\frac{\pi}{6}} \cos 	heta \ & =\frac{1}{3}[\sin 	heta]_0^{\frac{\pi}{6}} \ & I=\frac{1}{3}\left[\sin \frac{\pi}{6}-\sin 0ight] \ & I=\frac{1}{6} \ & 	herefore \quad \frac{k}{k+5}=\frac{1}{6} \ & 6 k=k+5 \ & 	herefore \quad \mathrm{k}=1 \ & \end{aligned}$

$\int_1^2 \frac{\mathrm{d} x}{\left(x^2-2 x+4\right)^{\frac{3}{2}}}=\frac{\mathrm{k}}{\mathrm{k}+5}$, then…

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