Evaluate: ∫_3^4▒(x^2+x+3)/(3x^5 ) dx
=[(x^3/3+x^2/2+3x)/(x^6/2)]
=(4^3/3+4^2/2+3(4))/(4^6/2)-(3^3/3+3^2/2+3(3))/(3^6/2)
=31/1536-5/81
=-1723/41472
I need some help with the working for this definite integral question. Iv shown what I have done but my answer is still not correct.
It is not true that $\int \frac f g=\frac {\int f} {\int g}$. So your very first step is wrong. Instead of this write $\frac {x^{2}+x+3} {3x^{5}}$ as $\frac 1 3 x^{-3}+\frac 1 3x^{-4}+x^{-5}$ and integrate all three terms.
$$\int_3^4 \frac{x^2+x+3}{3x^5}dx = \int_3^4 \left( \frac{x^2}{3x^5}+\frac{x}{3x^5}+\frac{3}{3x^5} \right) dx = \int_3^4 \left( \frac{1}{3}x^{-3}+\frac{1}{3}x^{-4}+x^{-5} \right) dx$$ Now, integrate each of this terms using the formula $$\int x^m dx = \frac{1}{m+1} x^{m+1}$$ and evaluate the final result on $4$ and substract the evaluation on $3$.
