Question

Integrate x12_,
30
dre
34-2​

Answer 1

Answer:

your answer

Step-by-step explanation:

∫x3x+1dx∫x3x+1dx

∫x3x+1dx∫x3x+1dx=∫x3+1−1x+1dx=∫x3+1−1x+1dx

∫x3x+1dx∫x3x+1dx=∫x3+1−1x+1dx=∫x3+1−1x+1dx=∫x3+1x+1dx−∫1x+1dx=∫x3+1x+1dx−∫1x+1dx

∫x3x+1dx∫x3x+1dx=∫x3+1−1x+1dx=∫x3+1−1x+1dx=∫x3+1x+1dx−∫1x+1dx=∫x3+1x+1dx−∫1x+1dx=∫(x2+1−x)dx−∫1x+1dx=∫(x2+1−x)dx−∫1x+1dx

∫x3x+1dx∫x3x+1dx=∫x3+1−1x+1dx=∫x3+1−1x+1dx=∫x3+1x+1dx−∫1x+1dx=∫x3+1x+1dx−∫1x+1dx=∫(x2+1−x)dx−∫1x+1dx=∫(x2+1−x)dx−∫1x+1dx=x33+x−x22+ln(x+1)+C