x-2-1-x-1-2-3-dx-

Question Number 129794 by bramlexs22 last updated on 19/Jan/21

\int (x^{2} - 1) {(x + 1)}^{- 2 / 3} dx ?

Answered by EDWIN88 last updated on 19/Jan/21

\int (x - 1) {(x + 1)}^{1 / 3} dx = (x - 1) (\frac{3}{4} {(x + 1)}^{4 / 3}) - \frac{3}{4} \int {(x + 1)}^{4 / 3} dx

= \frac{3}{4} (x - 1) {(x + 1)}^{4 / 3} - \frac{9}{28} {(x + 1)}^{7 / 3} + C

Answered by Lordose last updated on 19/Jan/21

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

I \overset{u = x + 1}{=} \int u^{\frac{1}{3}} (u - 2) du = \int u^{\frac{4}{3}} du - 2 \int u^{\frac{1}{3}} du

I = \frac{{3 u}^{\frac{7}{3}}}{7} - \frac{{3 u}^{\frac{4}{3}}}{2} + C

Commented by MJS_new last updated on 19/Jan/21

\dots = \frac{3}{14} u^{4 / 3} (2 u + 7) = \frac{3}{14} {(x + 1)}^{4 / 3} (2 x + 5) + C

Answered by Dwaipayan Shikari last updated on 19/Jan/21

\int (x - 1) {(x + 1)}^{\frac{1}{3}} d x

= \int (t - 2) t^{\frac{1}{3}} d t

= \frac{3}{7} t^{\frac{7}{3}} - \frac{3}{2} t^{\frac{4}{3}} d t = \frac{3}{7} {(x + 1)}^{\frac{7}{3}} - \frac{3}{2} {(x + 1)}^{\frac{4}{3}} + C