What is the integral of (4x^2+1)^4+(4x^2+1)^3 ?

The integral of (4x^2+1)^4+(4x^2+1)^3 is 256/9 x^9+320/7 x^7+144/5 x^5+28/3 x^3+2x+C

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integral of (4x^2+1)^4+(4x^2+1)^3

\int (4 x^{2} + 1)^{4} + (4 x^{2} + 1)^{3} d x

\frac{256}{9} x^{9} + \frac{320}{7} x^{7} + \frac{144}{5} x^{5} + \frac{28}{3} x^{3} + 2 x + C

Solution steps

\int (4 x^{2} + 1)^{4} + (4 x^{2} + 1)^{3} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int (4 x^{2} + 1)^{4} d x + \int (4 x^{2} + 1)^{3} d x

\int (4 x^{2} + 1)^{4} d x = \frac{1}{315} x (8960 x^{8} + 11520 x^{6} + 6048 x^{4} + 1680 x^{2} + 315)

\int (4 x^{2} + 1)^{3} d x = \frac{1}{35} x (320 x^{6} + 336 x^{4} + 140 x^{2} + 35)

= \frac{1}{315} x (8960 x^{8} + 11520 x^{6} + 6048 x^{4} + 1680 x^{2} + 315) + \frac{1}{35} x (320 x^{6} + 336 x^{4} + 140 x^{2} + 35)

Simplify

\frac{1}{315} x (8960 x^{8} + 11520 x^{6} + 6048 x^{4} + 1680 x^{2} + 315) + \frac{1}{35} x (320 x^{6} + 336 x^{4} + 140 x^{2} + 35) : \frac{256}{9} x^{9} + \frac{320}{7} x^{7} + \frac{144}{5} x^{5} + \frac{28}{3} x^{3} + 2 x

= \frac{256}{9} x^{9} + \frac{320}{7} x^{7} + \frac{144}{5} x^{5} + \frac{28}{3} x^{3} + 2 x

Add a constant to the solution

= \frac{256}{9} x^{9} + \frac{320}{7} x^{7} + \frac{144}{5} x^{5} + \frac{28}{3} x^{3} + 2 x + C

What is the integral of (4x^2+1)^4+(4x^2+1)^3 ?
The integral of (4x^2+1)^4+(4x^2+1)^3 is 256/9 x^9+320/7 x^7+144/5 x^5+28/3 x^3+2x+C