Question

integrate (1-x³)² with respect to x​

Answer 1

Answer:

Start off by expanding the square:

x^6 -2x³ + 1.

Now, we can integrate it in dx:

∫x^6 - 2x^3 + 1 dx = ∫x^6 dx - 2∫x^3 dx + ∫1 dx = (x^7)/7 - 2((x^4)/4) + x = (x^7)/7 - (x^4)/2 + x.

That's the result!

Answer 2

$\\ \sf\longmapsto {\displaystyle{\int}}(1-x^3)^2dx$

$\\ \sf\longmapsto {\displaystyle{\int}}(1-x^6+2x^3)dx$

$\\ \sf\longmapsto x-\dfrac{x^{6+1}}{6+1}+2\dfrac{x^{3+1}}{3+1}$

$\\ \sf\longmapsto x-\dfrac{x^7}{7}+2\dfrac{x^4}{4}$

$\\ \sf\longmapsto x-\dfrac{x^7}{7}+\dfrac{x^4}{2}$