To find the indefinite integral, we can use the reverse power rule: we raise the exponent by one and then divide by our new exponent.

\(\displaystyle \int 2x^4+\frac{1}{2}x{\mathrm{d} x}=\frac{2x^{4+1}}{4+1}+\frac{\frac{1}{2}x^{1+1}}{1+1}+c\)

Remember when taking the indefinite integral to include a \(\displaystyle +c\) to cover any potential constants.

Simplify.

\(\displaystyle \int 2x^4+\frac{1}{2}x{\mathrm{d} x}=\frac{2x^{5}}{5}+\frac{\frac{1}{2}x^{2}}{2}+c\)

\(\displaystyle \int 2x^4+\frac{1}{2}x{\mathrm{d} x}=\frac{2}{5}x^5+\frac{1}{4}x^{2}+c\)

To find the indefinite integral, we can use the reverse power rule: we raise the exponent by one and then divide by our new exponent.

We are going to treat \(\displaystyle 12\) as \(\displaystyle 12x^0\) since anything to the zero power is one.

\(\displaystyle \int 3x+12{\mathrm{d} x}=\frac{3x^{1+1}}{1+1}+\frac{12x^{0+1}}{0+1}+c\)

Remember when taking the indefinite integral to include a \(\displaystyle +c\) to cover any potential constants.

Simplify.

\(\displaystyle \int 3x+12{\mathrm{d} x}=\frac{3x^{2}}{2}+\frac{12x^{1}}{1}+c\)

\(\displaystyle \int 3x+12{\mathrm{d} x}=\frac{3}{2}x^2+12x+c\)

To find the indefinite integral, we use the reverse power rule. That means we raise the exponent on the variables by one and then divide by the new exponent.

\(\displaystyle \int 5x^2+12x{\mathrm{d} x}=\frac{5x^{2+1}}{2+1}+\frac{12x^{1+1}}{1+1}+c\)

Remember to include a \(\displaystyle +c\) when computing integrals. This is a place holder for any constant that might be in the new expression.

\(\displaystyle \int 5x^2+12x{\mathrm{d} x}=\frac{5x^{3}}{3}+\frac{12x^{2}}{2}+c\)

\(\displaystyle \int 5x^2+12x{\mathrm{d} x}=\frac{5}{3}x^3+\frac{12}{2}x^2+c\)

\(\displaystyle \int 5x^2+12x{\mathrm{d} x}=\frac{5}{3}x^3+6x^2+c\)

To find the indefinite integral, we use the reverse power rule. That means we raise the exponent on the variables by one and then divide by the new exponent.

\(\displaystyle \int -8x^2+15{\mathrm{d} x}=\frac{-8x^{2+1}}{2+1}+\frac{15x^{0+1}}{0+1}+c\)

Remember to include a \(\displaystyle +c\) when doing integrals. This is a placeholder for any constant that might be in the new expression.

\(\displaystyle \int -8x^2+15{\mathrm{d} x}=\frac{-8x^{3}}{3}+\frac{15x^{1}}{1}+c\)

\(\displaystyle \int -8x^2+15{\mathrm{d} x}=-\frac{8}{3}x^3+15x+c\)

To find the indefinite integral, we can use the reverse power rule. Raise the exponent of the variable by one and then divide by that new exponent.

We're going to treat \(\displaystyle 5\) as \(\displaystyle 5x^0\).

\(\displaystyle \int x^3+2x^2+5{\mathrm{d} x}=\frac{x^{3+1}}{3+1}+\frac{2x^{2+1}}{2+1}+\frac{5x^{0+1}}{0+1}+c\)

Remember to include the \(\displaystyle +c\) when taking the integral to compensate for any constant.

\(\displaystyle \int x^3+2x^2+5{\mathrm{d} x}=\frac{x^{4}}{4}+\frac{2x^{3}}{3}+\frac{5x^{1}}{1}+c\)

Simplify.

\(\displaystyle \int x^3+2x^2+5{\mathrm{d} x}=\frac{1}{4}x^4+\frac{2}{3}x^3+5x+c\)

To find the indefinite integral, we can use the reverse power rule. We raise the exponent of the variable by one and divide by our new exponent.

\(\displaystyle \int 4x^2+\frac{2}{3}x{\mathrm{d} x}=\frac{4x^{2+1}}{2+1}+\frac{\frac{2}{3}x^{1+1}}{1+1}+c\)

Remember to include a \(\displaystyle +c\) to cover any potential constant that might be in our new equation.

\(\displaystyle \int 4x^2+\frac{2}{3}x{\mathrm{d} x}=\frac{4x^{3}}{3}+\frac{\frac{2}{3}x^{2}}{2}+c\)

\(\displaystyle \int 4x^2+\frac{2}{3}x{\mathrm{d} x}=\frac{4}{3}x^3+\frac{1}{3}x^2+c\)

Just like with the derivatives, the indefinite integrals or anti-derivatives of trig functions must be memorized.

\(\displaystyle \int \sin(x){\mathrm{d} x}=-\cos(x)+c\)

To find the indefinite integral of our given equation, we can use the reverse power rule: we raise the exponent by one and then divide by that new exponent.

\(\displaystyle \int 4x^2+5x-3\: dx=\frac{4x^{2+1}}{2+1}+\frac{5x^{1+1}}{1+1}-\frac{3x^{0+1}}{0+1}+c\)

Don't forget to include a \(\displaystyle +c\) to compensate for any constant!

\(\displaystyle \int 4x^2+5x-3\: dx=\frac{4x^{3}}{3}+\frac{5x^{2}}{2}-\frac{3x^{1}}{1}+c\)

\(\displaystyle \int 4x^2+5x-3\: dx=\frac{4}{3}x^3+\frac{5}{2}x^2-3x+c\)

To find the indefinite integral, we're going to use the reverse power rule: raise the exponent of the variable by one and then divide by that new exponent.

\(\displaystyle \int 4x+3 dx=(\frac{4x^{1+1}}{1+1})+(\frac{3x^{0+1}}{0+1})+c\)

Be sure to include \(\displaystyle + c\) to compensate for any constant!

\(\displaystyle \int 4x+3dx=(\frac{4x^{1+1}}{1+1})+(\frac{3x^{0+1}}{0+1})+c\)

\(\displaystyle \int 4x+3dx=(\frac{4x^{2}}{2})+(\frac{3x^{1}}{1})+c\)

\(\displaystyle \int 4x+3dx=2x^2+3x+c\)

To find the indefinite integral, or anti-derivative, we can use the reverse power rule. We raise the exponent of each variable by one and divide by that new exponent.

\(\displaystyle \int x^4-3x^2+8\ {\mathrm{d} x}=(\frac{x^{4+1}}{4+1})-(\frac{3x^{2+1}}{2+1})+(\frac{8x^{0+1}}{0+1})+c\)

Don't forget to include a \(\displaystyle +c\) to cover any constant!

Simplify.

\(\displaystyle \int x^4-3x^2+8\ {\mathrm{d} x}=(\frac{x^{5}}{5})-(\frac{3x^{3}}{3})+(\frac{8x^{1}}{1})+c\)

\(\displaystyle \int x^4-3x^2+8\ {\mathrm{d} x}=\frac{1}{5}x^5-x^3+8x+c\)