calc

[heather chisholm]

find the integral
(1-6x)e(3x-9x^2)dx

 

[heather chisholm]

find the integral
7x^6dx/(4+x^7)^4

 

[soroban]

Hello, Heather chisholm!

More substitution . . .

\(\displaystyle \L\:\int (1\,-\,6x)\cdot e^{(3x-9x^2)}\,dx\)

Let \(\displaystyle u\,=\,3x-9x^2\;\;\Rightarrow\;\;du\,=\,(3-18x)\,dx\;\;\Rightarrow\;\;dx = \frac{du}{3(1-6x)}\)

Substitute: \(\displaystyle \L\;\int(1-6x)e^u\,\left(\frac{du}{3(1-6x)}\right)\;= \;\frac{1}{3}\int e^u\,du\;=\;\frac{1}{3}e^u\,+\,C\)

Back-substitute: \(\displaystyle \L\;\frac{1}{3}e^{(3x-9x^2)}\,+\,C\)

 

[soroban]

Hello, Heather!

I hope you're getting the idea of Substitution . . .

\(\displaystyle \L\int\frac{7x^6}{(4\,+\,x^7)^4}\,dx\)

Let \(\displaystyle u\,=\,4+x^7\;\;\Rightarrow\;\;dx\,=\,7x^6\,dx\;\;\Rightarrow\;\;dx\,=\,\frac{du}{7x^6}\)

Substitute: \(\displaystyle \L\;\int\frac{7x^6}{u^4}\,\left(\frac{du}{7x^6}\right) \;=\;\int u^{-4}du\;=\;\frac{u^{-3}}{-3}\,+\,C\;=\;-\frac{1}{3u^3}\,+\,C\)

Back-substitute: \(\displaystyle \L\;-\frac{1}{3(4\,+\,x^7)^3}\,+\,C\)