∫(x3+1)5x2dx\int \left(x^3+1\right)^5x^2dx∫(x3+1)5x2dx
Here, we apply substitution method I.e.
Let u=x3+1u = x^3 + 1u=x3+1
⟹ dudx=3x2\implies \frac{du}{dx} = 3x^2⟹dxdu=3x2
⟹ dx=13x2du\implies dx= \frac{1}{3x^2}du⟹dx=3x21du
Replacing back, we have;
∫(u)5x213x2du\int (u)^5 \cancel{x^2} \frac{1}{3 \cancel{x^2} }du∫(u)5x23x21du =13∫u5du= \frac{1}{3} \int u^5 du=31∫u5du
Now we solve;
∫u5du\int u^5 du∫u5du
Here, we apply power rule;
Where we let;
∫undu=un+1n+1\int u^n du = \frac{u^{n+1}}{n+1}∫undu=n+1un+1 With n=5n = 5n=5 , we have;
=u66= \frac{u^6}{6}=6u6
Replacing this back to 13∫u5du\frac{1}{3} \int u^5 du31∫u5du we have;
u618\frac{u^6}{18}18u6
Now we undo the substitution
u=x3+1u = x^3 +1u=x3+1
(x3+1)618\frac{(x^3 +1)^6}{18}18(x3+1)6 and we add a C to this, hence the final answer i.e.
(x3+1)618+C\frac{(x^3 +1)^6}{18} + C18(x3+1)6+C
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