What is the integral of (x^2+1)nx ?

The integral of (x^2+1)nx is n((x^4)/4+(x^2)/2)+C

Popular >

integral of (x^2+1)nx

\int (x^{2} + 1) n x d x

n (\frac{x ^{4}}{4} + \frac{x ^{2}}{2}) + C

Solution steps

\int (x^{2} + 1) n x d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= n \cdot \int (x^{2} + 1) x d x

Expand

(x^{2} + 1) x : x^{3} + x

= n \cdot \int x^{3} + x 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

= n (\int x^{3} d x + \int x d x)

\int x^{3} d x = \frac{x ^{4}}{4}

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

= n (\frac{x ^{4}}{4} + \frac{x ^{2}}{2})

Add a constant to the solution

= n (\frac{x ^{4}}{4} + \frac{x ^{2}}{2}) + C

\int_{8}^{R} \frac{1}{x ^{2} - 9 x + 18} d x

f (x) = \frac{6}{3 x + 1}, (- 1, - 3)

\int \frac{1}{2} ln (\frac{1 + x}{1 - x}) d x

\int \frac{6 x + 7}{( 9 x ^{2} + 21 x ) ^{4}} d x

n = 1 \sum \infty \frac{1 0 ^{n}}{9 ^{n}}

What is the integral of (x^2+1)nx ?
The integral of (x^2+1)nx is n((x^4)/4+(x^2)/2)+C