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#### paulmdrdo

##### Active member

- May 13, 2013

- 386

how would i go about solving these problems?

\begin{align*}\displaystyle \int\frac{xe^x}{(1+x)^2}dx\end{align*}

\begin{align*}\int\frac{(1-x)dx}{\sqrt{1-x^2}}\end{align*}

this is my solution to prob 2

\begin{align*}\displaystyle\int\frac{(1-x)dx}{\sqrt{1-x^2}}\,=\,\int\frac{dx}{\sqrt{1-x^2}}-\int\frac{x}{\sqrt{1-x^2}}dx\end{align*}

then,

\begin{align*}\displaystyle \sin^{-1}x+C_1-\int\frac{x}{\sqrt{1-x^2}}dx\\\\u\,=\,1-x^2\\du\,=\,-2xdx\\dx\,=\,\frac{du}{-2x}\\\\\int\frac{x}{\sqrt{u}(-2x)}du\\\\-\frac{1}{2}\int\frac{du}{\sqrt{u}}=\,-\frac{1}{2}\frac{(u^{-\frac{1}{2}})}{\frac{1}{2}}\\\\ \sin^{-1}x+(1-x^2)^{\frac{1}{2}}+C\end{align*}

is my answer correct in prob 2?

\begin{align*}\displaystyle \int\frac{xe^x}{(1+x)^2}dx\end{align*}

\begin{align*}\int\frac{(1-x)dx}{\sqrt{1-x^2}}\end{align*}

this is my solution to prob 2

\begin{align*}\displaystyle\int\frac{(1-x)dx}{\sqrt{1-x^2}}\,=\,\int\frac{dx}{\sqrt{1-x^2}}-\int\frac{x}{\sqrt{1-x^2}}dx\end{align*}

then,

\begin{align*}\displaystyle \sin^{-1}x+C_1-\int\frac{x}{\sqrt{1-x^2}}dx\\\\u\,=\,1-x^2\\du\,=\,-2xdx\\dx\,=\,\frac{du}{-2x}\\\\\int\frac{x}{\sqrt{u}(-2x)}du\\\\-\frac{1}{2}\int\frac{du}{\sqrt{u}}=\,-\frac{1}{2}\frac{(u^{-\frac{1}{2}})}{\frac{1}{2}}\\\\ \sin^{-1}x+(1-x^2)^{\frac{1}{2}}+C\end{align*}

is my answer correct in prob 2?

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