How do you evaluate \int x ( x ^ { 2} + 1) ^ { 8} d x?

Nov 7, 2017

$\therefore \int x {\left({x}^{2} + 1\right)}^{8} \mathrm{dx} = \frac{1}{18} {\left({x}^{2} + 1\right)}^{9} + C$

Explanation:

$\int x {\left({x}^{2} + 1\right)}^{8} \mathrm{dx}$

remember integration is the reverse operation of differentiation.

1) Inspection

we note that the function outside the bracket is the bracket differentiated, mutliplied by a constant.

so we suspect that the integral is the bracket tot eh power $8 + 1$

lets try it

d/(dx)((x^+1)^9=9xx2x(x^2+1)^8=18x(x^2+1)^8

which is the integrand except for the $18$

$\therefore \int x {\left({x}^{2} + 1\right)}^{8} \mathrm{dx} = \frac{1}{18} {\left({x}^{2} + 1\right)}^{9} + C$

#2) substitution

$\int x {\left({x}^{2} + 1\right)}^{8} \mathrm{dx}$

$u = {x}^{2} + 1 \implies \mathrm{du} = 2 x \mathrm{dx}$

$\int x {\left({x}^{2} + 1\right)}^{8} \mathrm{dx} = \int \cancel{x} {u}^{8} \times \frac{\mathrm{du}}{2 \cancel{x}}$

$= \frac{1}{2} \times \frac{1}{9} {u}^{9} + c$

$= \frac{1}{18} {u}^{9} + C = \frac{1}{18} {\left({x}^{2} + 1\right)}^{9} + C$