What is the integral of (5x^4)/4+(3x^2)/2+7x+C ?

The integral of (5x^4)/4+(3x^2)/2+7x+C is (x^5)/4+(x^3)/2+(7x^2)/2+Cx+C

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integral of (5x^4)/4+(3x^2)/2+7x+C

\int \frac{5 x ^{4}}{4} + \frac{3 x ^{2}}{2} + 7 x + C d x

\frac{x ^{5}}{4} + \frac{x ^{3}}{2} + \frac{7 x ^{2}}{2} + C x + C

Solution steps

\int \frac{5 x ^{4}}{4} + \frac{3 x ^{2}}{2} + 7 x + C d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{5 x ^{4}}{4} d x + \int \frac{3 x ^{2}}{2} d x + \int 7 x d x + \int C d x

\int \frac{5 x ^{4}}{4} d x = \frac{x ^{5}}{4}

\int \frac{3 x ^{2}}{2} d x = \frac{x ^{3}}{2}

\int 7 x d x = \frac{7 x ^{2}}{2}

\int C d x = C x

= \frac{x ^{5}}{4} + \frac{x ^{3}}{2} + \frac{7 x ^{2}}{2} + C x

Add a constant to the solution

= \frac{x ^{5}}{4} + \frac{x ^{3}}{2} + \frac{7 x ^{2}}{2} + C x + C

\frac{s + 1}{( s + 1 ) ^{2} + 1}

\int (11 x)^{2} d x

\int \frac{1}{( x + 4 ) ^{\frac{7}{2}}} d x

y^{^{'}} + 2 x y = 1

(- 2)^{^{'}}

What is the integral of (5x^4)/4+(3x^2)/2+7x+C ?
The integral of (5x^4)/4+(3x^2)/2+7x+C is (x^5)/4+(x^3)/2+(7x^2)/2+Cx+C