2
$\begingroup$

I am trying to find an indefinite integral. The question suggests that it can be solved with integration by substitution, but I cannot see how. Multiplying out the brackets and integrating gives an eight-order result. Can anyone help here?

$$ \int \left(x+4\right)\left(\frac{1}{3}x+8\right)^6\:dx $$

$\endgroup$
  • 2
    $\begingroup$ Multiply and divide by $3^6$ (the one in the numerator goes into the exponent expression to cancel out the $\frac 13$). Next, put $y=x+24$. Now see what you get. $\endgroup$ – астон вілла олоф мэллбэрг Mar 6 '17 at 11:12
2
$\begingroup$

Isn't it natural to set $u=x/3+8$ ?

Then

$$\int \left(x+4\right)\left(\frac{1}{3}x+8\right)^6\:dx=3\int \left(3u-20\right)u^6\:du$$

which should be easy. (By the way, in all cases integrating a polynomial is easy.)

$\endgroup$
  • $\begingroup$ Yes, thanks very much! The 3 factor confused me for a while, until I realised you're substituting dx = 3 du. $\endgroup$ – Simon Mar 7 '17 at 11:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.