∫ ( x x + 2 4 ) d x u = x + 2 − − > y − 2 = x d u = 1 d x {\displaystyle \int _{}^{}\left({\frac {x}{\sqrt[{4}]{x+2}}}\right)dxu=x+2-->y-2=xdu=1dx}
![{\displaystyle \int _{}^{}\left({\frac {x}{\sqrt[{4}]{x+2}}}\right)dxu=x+2-->y-2=xdu=1dx}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/c82831324555d0d77555e4074a9ee861e562b413)