Вычислить длину дуги, буду благодарна

Решите задачу:

$L=\int_{2}^{6}\sqrt{1+\left (\left (\sqrt{(x-2)^3} \right )' \right )^2}\mathrm{dx}=\int_{2}^{6}\sqrt{1+{9\over4}\cdot(x-2)}\mathrm{dx}=\int_{2}^{6}\sqrt{{9\over4}x-{7\over2}}\mathrm{dx}={4\over9}\int_{2}^{6}\sqrt{{9\over4}x-{7\over2}}\; \; \mathrm{d\left ({9\over4}x-{7\over2} \right )}={4\over9}\cdot\sqrt{\left ({9\over4}x-{7\over2} \right )^3}\cdot{2\over3}|_{2}^{6}={8\over27}\cdot\sqrt{\left ({9\over4}x-{7\over2} \right )^3}|_{2}^{6}={8\over27}\cdot\left ( \sqrt{\left ({9\over4}\cdot6-{7\over2} \right )^3}-\sqrt{\left ({9\over4}\cdot2-{7\over2} \right )^3} \right )={8\over27}\cdot\left ( 10\sqrt{10}-1 \right )\approx 9,07342$