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Show, using the following methods, that the indefinite integral of x3/(x+1)1/2x^3 /(x+1)^{1 / 2} is

J=235(5x36x2+8x16)(x+1)1/2+c.J=\frac{2}{35}\left(5 x^3-6 x^2+8 x-16\right)(x+1)^{1 / 2}+c .

(a) Repeated integration by parts. (b) Setting x+1=u2x+1=u^2 and determining dJ/dud J / d u as (dJ/dx)(dx/du)(d J / d x)(d x / d u).

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Answered 2 years ago
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(a):x3(x+1)12dx\textbf{(a)}:\int x^{3}\left(x+1 \right)^{\frac{-1}{2}}dx

J=x3(x+1)12dx=x3(x+1)12123x2(x+1)1212dx=2x3(x+1)126x2(x+1)12dx=2x3(x+1)126[x2(x+1)3232]2x(x+1)3232dx=2x3(x+1)124x2(x+1)32+8x(x+1)32dx=2x3(x+1)124x2(x+1)32+8x(x+1)52522x3(x+1)124x2(x+1)5252dx=2x3(x+1)124x2(x+1)32+165x(x+1)52165(x+1)52dx=2x3(x+1)124x2(x+1)32+165x(x+1)5216527(x+1)72+C\begin{align*} &J= \int x^{3}\left(x+1 \right)^{\frac{-1}{2}} dx\\\\ &=\dfrac{x^{3}\left(x+1 \right)^{\frac{1}{2}}}{\dfrac{1}{2}}- \int \dfrac{3x^{2}\left(x+1 \right)^{\frac{1}{2}}}{\dfrac{1}{2}} dx\\\\ &=2x^{3}\left(x+1 \right)^{\frac{1}{2}}- 6\int x^{2}\left(x+1 \right)^{\frac{1}{2}}dx\\\\ &=2x^{3}\left(x+1 \right)^{\frac{1}{2}}- 6\left[\dfrac{x^{2}\left(x+1 \right)^{\frac{3}{2}}}{\dfrac{3}{2}} \right]- \int \dfrac{2x\left(x+1 \right)^{\frac{3}{2}}}{\dfrac{3}{2}} dx\\\\ &=2x^{3}\left(x+1 \right)^{\frac{1}{2}}- 4x^{2}\left(x+1 \right)^{\frac{3}{2}}+8\int x\left(x+1 \right)^{\frac{3}{2}} dx\\\\ &=2x^{3}\left(x+1 \right)^{\frac{1}{2}}- 4x^{2}\left(x+1 \right)^{\frac{3}{2}}+8\dfrac{x\left(x+1 \right)^{\frac{5}{2}}}{\dfrac{5}{2}}- \int \dfrac{2x^{3}\left(x+1 \right)^{\frac{1}{2}}- 4x^{2}\left(x+1 \right)^{\frac{5}{2}}}{\dfrac{5}{2}} dx\\\\ &= 2x^{3}\left(x+1 \right)^{\frac{1}{2}}- 4x^{2}\left(x+1 \right)^{\frac{3}{2}}+\dfrac{16}{5} x\left(x+1 \right)^{\frac{5}{2}}- \dfrac{16}{5} \int \left(x+1 \right)^{\frac{5}{2}}dx\\\\ &=2x^{3}\left(x+1 \right)^{\frac{1}{2}}- 4x^{2}\left(x+1 \right)^{\frac{3}{2}}+\dfrac{16}{5} x\left(x+1 \right)^{\frac{5}{2}}- \dfrac{16}{5}\cdot \dfrac{2}{7}\left(x+1 \right)^{\frac{7}{2}}+C\\\\ \end{align*}

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