Question

$\int x^{27}\left(6x^2+5x+4\right){\left(1+x+x^2\right)}^{6}dx$ is equal to

• A. $\frac{x^4}{7}\cdot {\left(1+x+x^2\right)}^7$
• B. $\frac{x^{28}{\left(1+x+x^2\right)}^7}{7}+C$
• C. $\frac{x^{28}{\left(1+x+x^2\right)}^7}{28}+C$
• D. $\frac{x^{28}{\left(1+x+x^2\right)}^6}{6}+C$

Answer 1

Answer: (B)

$I=\int \left(6x^5+5x^4+4x^3\right){\left(x^4+x^5+x^6\right)}^{6}dx$
Let $x^6+x^5+x^4=t$
$\left(6x^5+5x^4+4x^2\right)dx=dt$
$I=\int t^{6}dt=\frac{{\left(x^4+x^5+x^6\right)}^7}{7}+C=\frac{x^{28}{\left(1+x+x^2\right)}^7}{7}+C$