Please check my work and lend a helping hand

[shivers20]

I did not know how to post images. Please copy and paste the link. Please check my work. Thanks in advance. [/code]

 

[jonboy]

It is impossible to fix those links. Re-upload your pic and copy and paste the "Direct link to image" at the bottom of Imageshack. Then we will help you.

 

[shivers20]

It is impossible to fix those links. Re-upload your pic and copy and paste the "Direct link to image" at the bottom of Imageshack. Then we will help you.

 

[soroban]

Hello, shivers20!

\(\displaystyle \L1)\:\int(x^3\,+\,5x\,-\,7)\,dx \;=\;\frac{x^4}{4} + \frac{5x^2}{2}\,-\,7x\) + C

\(\displaystyle \L2)\;\int\left(x^{\frac{1}{2}}\,+\,4x^{-3}\right)\,dx\)

\(\displaystyle = \;\frac{x^{\frac{3}{2}}}{\frac{3}{2}} \,+\,\frac{4x^{^{-2}}}{-2}\,+\,C\;=\;\frac{2}{3}x^{\frac{3}{2}} \,-\,2x^{^{-2}}\,+\,C\)

\(\displaystyle \L3)\;\int\frac{r}{\sqrt{1\,+\,r^2}}\,dr\)

Let \(\displaystyle u\:=\:1\,+\,r^2\;\;\Rightarrow\;\;du\,=\,2r\,dr\;\;\Rightarrow\;\;dr\,=\,\frac{dt}{2r}\)

Substitute: \(\displaystyle \L\,\int\frac{r}{t^{\frac{1}{2}}}\,\frac{dt}{2r}\;=\;\frac{1}{2}\int t^{-\frac{1}{2}}\,dt\;=\;\frac{1}{2}\frac{t^{\frac{1}{2}}}{\frac{1}{2}}\,+\,C \;= \;t^{\frac{1}{2}}\,+\,C\)

Back-substitute:\(\displaystyle \,(1\,+\,r^2)^{\frac{1}{2}}\,+\,C\)

\(\displaystyle \L 4)\;\int(2\,-\,3x)^{\frac{3}{5}}dx\)

Let \(\displaystyle u\:=\:2\,-\,3x\;\;\Rightarrow\;\;du\,=\,-3\,dx\;\;\Rightarrow\;\;dx\,=\,-\frac{du}{3}\)

Substitute: \(\displaystyle \L\,\int u^{\frac{3}{5}}\left(-\frac{du}{3}\right) \;= \;-\frac{1}{3}\int u^{\frac{3}{5}}du \;= \;-\frac{1}{3}\frac{u^{\frac{8}{5}}}{\frac{8}{5}}\,+\,C \;=\;-\frac{5}{24}u^{\frac{8}{5}}\,+\,C\)

Back-substitute: \(\displaystyle \L\:-\frac{5}{24}(2\,-\,3x)^{\frac{8}{5}}\,+\,C\)
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