Concepcion Hale

2022-01-07

Given $\int \left({x}^{5}-3{x}^{4}+6{x}^{3}+3\right)dx$, evaluate the indefinite integral.

Beverly Smith

Step 1
The given integral is $\int \left({x}^{5}-3{x}^{4}+6{x}^{3}+3\right)dx$.
obtain the integral value as follows.
$\int \left({x}^{5}-3{x}^{4}+6{x}^{3}+3\right)dx=\int {x}^{5}dx-\int 3{x}^{4}dx+\int 6{x}^{3}dx+\int 3dx$
$=\frac{{x}^{6}}{6}-3\left(\frac{{x}^{5}}{5}\right)+6\left(\frac{{x}^{4}}{4}\right)+3x+C$
$=\frac{{x}^{6}}{6}-\frac{3{x}^{5}}{5}+\frac{3{x}^{4}}{2}+3x+C$
Thus, the value of the integral is $\frac{{x}^{6}}{6}-\frac{3{x}^{5}}{5}+\frac{3{x}^{4}}{2}+3x$
Step 2
The value of the given integral is $\left(\frac{{x}^{6}}{6}\right)-3\left(\frac{{x}^{5}}{5}\right)+3\left(\frac{{x}^{4}}{2}\right)+3x$.

ambarakaq8

Given integral.
$\int {x}^{5}-3{x}^{4}+6{x}^{3}+3dx$
$\int {x}^{5}dx-\int 3{x}^{4}dx+\int 6{x}^{3}dx+\int 3dx$
Evaluate
$\frac{{x}^{6}}{6}-\frac{3{x}^{5}}{5}+\frac{3{x}^{4}}{2}+3x$
$\frac{{x}^{6}}{6}-\frac{3{x}^{5}}{5}+\frac{3{x}^{4}}{2}+3x+C$

karton

$I=\int \left({x}^{5}-3{x}^{4}+6{x}^{3}+3\right)dx\phantom{\rule{0ex}{0ex}}\text{we know that}\phantom{\rule{0ex}{0ex}}\int {x}^{n}dx=\frac{{x}^{n+1}}{n+1}\phantom{\rule{0ex}{0ex}}I=\frac{{x}^{5+1}}{5+1}-\frac{3{x}^{4+1}}{4+1}+\frac{6{x}^{3+1}}{3+1}+3x\phantom{\rule{0ex}{0ex}}I=\frac{{x}^{6}}{6}-\frac{3{x}^{5}}{5}+\frac{6{x}^{4}}{4}+3x\phantom{\rule{0ex}{0ex}}I=\frac{{x}^{6}}{6}-\frac{3}{5}{x}^{5}+\frac{3}{2}{x}^{4}+3x$