Question

integral of (2x^2+4x+1)^2​

Answer 1

Step-by-step explanation:

We have,

$\int(2 {x}^{2} + 4x + 1)^{2} dx$

$= \int \{(2x + 1)^{2} \}^{2} dx$

$= \int (2x + 1)^{4}dx$

$= \int (2x)^{4} + (2x)^{3} + {(2x)}^{2} +( 2x) + 1 dx$

$= \int (2x)^{4} dx + \int(2x)^{3} dx+ \int{(2x)}^{2}dx + \int( 2x) dx+ \int dx$

$= 16\int (x)^{4} dx + 8\int \: (x)^{3} dx+ 4\int{(x)}^{2}dx + 2\int(x) dx+ \int dx$

$= 16\frac{ x^{5}}{5} + 8\frac{ x^{4}}{4} + 4\frac{{x}^{3}}{3} + 2\frac{x^{2} }{2} + x + C$

$= \frac{16 }{5}x^{5} + \frac{ 8}{4}x^{4}+ \frac{4}{3}{x}^{3} +\frac{2}{2}x^{2}+ x + C$

$= \frac{16 }{5}x^{5} + 2x^{4}+ \frac{4}{3}{x}^{3} + x^{2}+ x + C$