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

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

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

$$\mathrm{Synthax}\mathrm{error}!\mathrm{haha}!$$

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

$$posonst=\sqrt{3x+2}x=\frac{t^2-2}{3}dx=2tdt$$$$\int \frac{2x^2+3}{\sqrt{3x+2}}dx=\int \frac{2{(t^2-2)}^2+27}{9t}2tdt=\int \frac{4{(t^2-2)}^2+54}{9}dt$$$$=\int [\frac{4}{9}(t^4-4t^2)+\frac{70}{9}]dt]$$$$=\frac{4}{45}{t}^5-\frac{16}{27}{t}^3+\frac{70}{9}t+C$$$$\int \frac{2x^2+3}{\sqrt{3x+2}}=[\frac{4}{45}{(3x+2)}^2-\frac{16}{27}(3x+2)+\frac{70}{9}]\sqrt{3x+2}+C$$

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

$$I=\int \frac{2x^2+3}{\sqrt{3x+2}}dx=\int \frac{2x^2}{\sqrt{3x+2}}dx+\int \frac{3}{\sqrt{3x+2}}dx$$$$=\frac{2}{9}\int \frac{{(3x+2-2)}^2}{\sqrt{3x+2}}dx+2\sqrt{3x+2}$$$$=\frac{2}{9}\int \frac{{(3x+2)}^2-4(3x+2)+4}{\sqrt{3x+2}}dx+2\sqrt{3x+2}$$$$=\frac{2}{9}\int ({(3x+2)}^{\frac{3}{2}}-4{(3x+2)}^{\frac{1}{2}}+4{(3x+2)}^{-\frac{1}{2}})dx+2\sqrt{3x+2}$$$$=\frac{2}{27}(\frac{2}{5}{(3x+2)}^{\frac{5}{2}}-\frac{8}{3}{(3x+2)}^{\frac{3}{2}}+8{(3x+2)}^{\frac{1}{2}})+2{(3x+2)}^{\frac{1}{2}}+C$$$$=\frac{4}{135}{(3x+2)}^{\frac{5}{2}}-\frac{16}{81}{(3x+2)}^{\frac{3}{2}}+\frac{70}{27}{(3x+2)}^{\frac{1}{2}}+C$$

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

$$I=\int \frac{2x^2+3}{\sqrt{3x+2}}dx,t=3x+2\Rightarrow x=\frac{t-2}{3}$$$$=\frac{1}{3}\int \frac{2{\left(\frac{t-2}{3}\right)}^2+3}{\sqrt{t}}dt=\frac{1}{27}\int \frac{2t^2-8t+35}{\sqrt{t}}dt$$$$=\frac{1}{27}(\frac{4}{5}{t}^{\frac{5}{2}}-\frac{16}{3}{t}^{\frac{3}{2}}+70t^{\frac{1}{2}})+C$$

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