How do you find the antiderivative of $\int {(x^{2} + 1)}^{2} d x$ $int (x^2+1)^2 dx$ ?

Question

8600 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-22T11:39:43

$\int {(x^{2} + 1)}^{2} d x = \frac{x^{5}}{5} + 2 \frac{x^{3}}{3} + x + c$ $int(x^2+1)^2dx=x^5/5+2x^3/3+x+c$

$\int {(x^{2} + 1)}^{2} d x =$ $int(x^2+1)^2dx=$

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

$\frac{x^{5}}{5} + 2 \frac{x^{3}}{3} + x + c$ $x^5/5+2x^3/3+x+c$ , $c$ $c$ $\in$ $in$ $RR$

How do you find the antiderivative of int (x^2+1)^2 dx∫(x2+1)2dxint (x^2+1)^2 dx?