Question

Evaluate...............

Answer 1

confused_____!!!!!!!!!

Answer 2

✪ᴀɴsᴡᴇʀ✪

$\boxed{\int \frac{{x}^{-\frac{1}{2}}}{1+{x}^{\frac{1}{3}}}dx= 6[{x}^{\frac{1}{6}}-{tan}^{-1}( {x}^{\frac{1}{6}})] +C}$

★ᴇxᴘʟᴀɪɴᴀᴛɪᴏɴ★

$I = \int \frac{{x}^{-\frac{1}{2}}}{1+{x}^{\frac{1}{3}}}dx$

$I = \int \frac{dx}{{x}^{\frac{1}{2}}(1+{x}^{\frac{1}{3}}) }$

Substitute ${x}^{\frac{1}{3}} = tan²(θ)$

$⇒ x= tan⁶(θ)$

$⇒ dx= 6tan⁵(θ)sec²(θ)dθ$

$I = \int \frac{6tan⁵(θ)sec²(θ)dθ}{tan³(θ)(1+tan²(θ))}$

$I = \int \frac{6tan²(θ)sec²(θ)dθ}{sec²(θ)}$

$I = \int 6tan²(θ)dθ$

$I = 6\int (sec²(θ)-1)dθ$

$I = 6[tan(θ)-θ] +C$

$I = 6[{x}^{\frac{1}{6}}-{tan}^{-1}( {x}^{\frac{1}{6}})] +C$