Answer :
Answer:
Step-by-step explanation:
I=∫(x²+1)³dx
=∫[(x²)³+1³+3x²×1(x²+1)]dx
=∫(x^6+1+3x^4+3x²)dx
=∫(x^6+3x^4+3x²+1)dx
=(x^7)/7 +3(x^5)/5+3x³/3+x+c
[tex]=\frac{x^7}{7} +\frac{3}{5} x^5+x^3+x+c[/tex]
Answer:
Step-by-step explanation:
I=∫(x²+1)³dx
=∫[(x²)³+1³+3x²×1(x²+1)]dx
=∫(x^6+1+3x^4+3x²)dx
=∫(x^6+3x^4+3x²+1)dx
=(x^7)/7 +3(x^5)/5+3x³/3+x+c
[tex]=\frac{x^7}{7} +\frac{3}{5} x^5+x^3+x+c[/tex]