Answer
$\frac{(x^3+3x)^5}{15} +C$
Work Step by Step
Evaluate the Integral using substitution: $\int (x^2+1)(x^3+3x)^4dx$
Substitution Rule: $\int f(g(x))g’(x)dx = \int f(u)du$
$u= x^3+3x$
$du =3x^2 + 3$
Since $du$ in the expression is equal to $(x^2+1)$ it must be multiplied by $\frac{1}{3}$
Solve the integral in terms of $u$:
$\int (u)^4(\frac13)du$
$\frac{1}{3}\int (u)^4du $
$\frac{1}{3}\frac{u^5}{5} +C$
$\frac{u^5}{15} + C$
Substitute for $u$:
$\frac{(x^3+3x)^5}{15} +C$
