2x-2-3-3x-2-

Question Number 175436 by mathlove last updated on 30/Aug/22

\int \frac{2 x^{2} + 3}{\sqrt{3 x + 2}} = ?

Commented by Ar Brandon last updated on 30/Aug/22

Synthax error! haha!

Commented by a.lgnaoui last updated on 30/Aug/22

p o s o n s t = \sqrt{3 x + 2} x = \frac{t^{2} - 2}{3} d x = 2 t d t

\int \frac{2 x^{2} + 3}{\sqrt{3 x + 2}} d x = \int \frac{2 {(t^{2} - 2)}^{2} + 27}{9 t} 2 t d t = \int \frac{4 {(t^{2} - 2)}^{2} + 54}{9} d t

= \int [\frac{4}{9} (t^{4} - 4 t^{2}) + \frac{70}{9}] d t]

= \frac{4}{45} t^{5} - \frac{16}{27} t^{3} + \frac{70}{9} t + C

\int \frac{2 x^{2} + 3}{\sqrt{3 x + 2}} = [\frac{4}{45} {(3 x + 2)}^{2} - \frac{16}{27} (3 x + 2) + \frac{70}{9}] \sqrt{3 x + 2} + C

Answered by Ar Brandon last updated on 30/Aug/22

I = \int \frac{2 x^{2} + 3}{\sqrt{3 x + 2}} d x = \int \frac{2 x^{2}}{\sqrt{3 x + 2}} d x + \int \frac{3}{\sqrt{3 x + 2}} d x

= \frac{2}{9} \int \frac{{(3 x + 2 - 2)}^{2}}{\sqrt{3 x + 2}} d x + 2 \sqrt{3 x + 2}

= \frac{2}{9} \int \frac{{(3 x + 2)}^{2} - 4 (3 x + 2) + 4}{\sqrt{3 x + 2}} d x + 2 \sqrt{3 x + 2}

= \frac{2}{9} \int ({(3 x + 2)}^{\frac{3}{2}} - 4 {(3 x + 2)}^{\frac{1}{2}} + 4 {(3 x + 2)}^{- \frac{1}{2}}) d x + 2 \sqrt{3 x + 2}

= \frac{2}{27} (\frac{2}{5} {(3 x + 2)}^{\frac{5}{2}} - \frac{8}{3} {(3 x + 2)}^{\frac{3}{2}} + 8 {(3 x + 2)}^{\frac{1}{2}}) + 2 {(3 x + 2)}^{\frac{1}{2}} + C

= \frac{4}{135} {(3 x + 2)}^{\frac{5}{2}} - \frac{16}{81} {(3 x + 2)}^{\frac{3}{2}} + \frac{70}{27} {(3 x + 2)}^{\frac{1}{2}} + C

Answered by Ar Brandon last updated on 30/Aug/22

I = \int \frac{2 x^{2} + 3}{\sqrt{3 x + 2}} d x, t = 3 x + 2 \Rightarrow x = \frac{t - 2}{3}

= \frac{1}{3} \int \frac{2 {(\frac{t - 2}{3})}^{2} + 3}{\sqrt{t}} d t = \frac{1}{27} \int \frac{2 t^{2} - 8 t + 35}{\sqrt{t}} d t

= \frac{1}{27} (\frac{4}{5} t^{\frac{5}{2}} - \frac{16}{3} t^{\frac{3}{2}} + 70 t^{\frac{1}{2}}) + C