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# Show, using the following methods, that the indefinite integral of $x^3 /(x+1)^{1 / 2}$ is$J=\frac{2}{35}\left(5 x^3-6 x^2+8 x-16\right)(x+1)^{1 / 2}+c .$(a) Repeated integration by parts. (b) Setting $x+1=u^2$ and determining $d J / d u$ as $(d J / d x)(d x / d u)$.

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$\textbf{(a)}:\int x^{3}\left(x+1 \right)^{\frac{-1}{2}}dx$

\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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