What is the integral of x^4*n^{-1}*sqrt(1)-x^2 ?

The integral of x^4*n^{-1}*sqrt(1)-x^2 is (x^5}{5n}-\frac{x^3)/3+C

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integral of x^4n^{-1}sqrt(1)-x^2

integral

x^{4} \cdot n^{- 1} \cdot 1 - x^{2}

\frac{x ^{5}}{5 n} - \frac{x ^{3}}{3} + C

Solution steps

\int x^{4} n^{- 1} 1 - x^{2} d x

Simplify

x^{4} n^{- 1} 1 - x^{2} : \frac{x ^{4}}{n} - x^{2}

= \int \frac{x ^{4}}{n} - x^{2} 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{x ^{4}}{n} d x - \int x^{2} d x

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

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

= \frac{x ^{5}}{5 n} - \frac{x ^{3}}{3}

Add a constant to the solution

= \frac{x ^{5}}{5 n} - \frac{x ^{3}}{3} + C

y = 2 e^{4} x + 3

- 3 sec (x) \cdot sec^{3} (x)

- sec^{2} (x)

- 3 sec (x) \cdot (sec (x))^{3}

\frac{1}{( s + a )}

What is the integral of x^4*n^{-1}*sqrt(1)-x^2 ?
The integral of x^4*n^{-1}*sqrt(1)-x^2 is (x^5}{5n}-\frac{x^3)/3+C