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Question Number 146929 by phally last updated on 16/Jul/21

Commented by phally last updated on 16/Jul/21

$$\mathrm{help}\mathrm{me}?$$

Commented by tabata last updated on 16/Jul/21

$$A)let:y^6=2x+1\Rightarrow 3y^{5}dy=dx$$$$$$$$\left(A\right)=\int \frac{3y^5}{y^4-y^3}dy=3\int \frac{y^2-1+1}{y-1}dy$$$$$$$$\left(A\right)=3\int [(y+1)+\frac{1}{y-1}]dy=3\frac{y^2}{2}+3y+3ln\mid y-1\mid +C$$$$$$$$replaseyby\sqrt[6]{2x+1}$$$$$$$$\therefore \left(A\right)=\int \frac{dx}{\sqrt[3]{{(2x+1)}^2}-\sqrt{2x+1}}=\frac{3}{2}\sqrt[3]{2x+1}+3\sqrt[6]{2x+1}+3ln\mid \sqrt[6]{2x+1}-1\mid +C$$$$⟨M:T⟩$$$$$$

Commented by tabata last updated on 16/Jul/21

$$\left(B\right)let{w}^6=1-2x\Rightarrow -3w^{5}dw=dx$$$$$$$$\left(B\right)=\int \frac{-3w^{5}dw}{w^3+w^2}=-3\int \frac{w^3+1-1}{w+1}dw$$$$$$$$\left(B\right)=-3\int (w^2-w+1-\frac{1}{w+1})dw=-3(\frac{w^3}{3}-\frac{w^2}{2}+w-ln\mid w+1\mid )+C$$$$$$$$replasew=\sqrt[6]{1-2x}$$$$$$$$\therefore \left(B\right)=\int \frac{dx}{\sqrt{1-2x}+\sqrt[3]{1-2x}}=\frac{3}{2}\sqrt[3]{1-2x}-\sqrt{1-2x}-3\sqrt[6]{1-2x}+3ln\mid \sqrt[6]{1-2x}+1\mid +C$$$$$$$$⟨M.T⟩$$

Commented by phally last updated on 16/Jul/21

$$\mathrm{thank}\mathrm{youw}$$

Commented by tabata last updated on 16/Jul/21

$$youarewelcome$$

Answered by behi834171 last updated on 16/Jul/21

$$A.$$$$let:2\mathit{x}+1={\mathit{t}}^6\Rightarrow \mathit{t}={(2\mathit{x}+1)}^{\frac{1}{6}},dx=3t^{5}dt$$$$\Rightarrow A=\int \frac{3t^{5}dt}{t^4-t^3}=\int \frac{3t^{2}dt}{t-1}=\int \frac{3(t^2-1)+3}{t-1}dt=$$$$=\int 3[t+1+\frac{1}{t-1}]dt=$$$$=3[\frac{1}{2}{t}^2+t+ln\mid t-1\mid +const.=$$$$=\frac{3}{2}\sqrt[3]{2\mathit{x}+1}+3\sqrt[6]{2\mathit{x}+1}+3\mathit{l}\mathit{n}\mid [\sqrt[6]{2\mathit{x}+1}-1]\mid +\mathit{C}.◼$$

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