What is the integral of (x^2+1)(x^3+3x) ?

The integral of (x^2+1)(x^3+3x) is (x^6)/6+x^4+3/2 x^2+C

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integral of (x^2+1)(x^3+3x)

\int (x^{2} + 1) (x^{3} + 3 x) d x

\frac{x ^{6}}{6} + x^{4} + \frac{3}{2} x^{2} + C

Solution steps

\int (x^{2} + 1) (x^{3} + 3 x) d x

Apply u-substitution

: \int \frac{( u + 1 ) ( u + 3 )}{2} d u

= \int \frac{( u + 1 ) ( u + 3 )}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int (u + 1) (u + 3) d u

Expand

(u + 1) (u + 3) : u^{2} + 4 u + 3

= \frac{1}{2} \cdot \int u^{2} + 4 u + 3 d u

Apply the Sum Rule:

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

= \frac{1}{2} (\int u^{2} d u + \int 4 u d u + \int 3 d u)

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

\int 4 u d u = 2 u^{2}

\int 3 d u = 3 u

= \frac{1}{2} (\frac{u ^{3}}{3} + 2 u^{2} + 3 u)

Substitute back

u = x^{2}

= \frac{1}{2} (\frac{( x ^{2} ) ^{3}}{3} + 2 (x^{2})^{2} + 3 x^{2})

Simplify

\frac{1}{2} (\frac{( x ^{2} ) ^{3}}{3} + 2 (x^{2})^{2} + 3 x^{2}) : \frac{x ^{6}}{6} + x^{4} + \frac{3}{2} x^{2}

= \frac{x ^{6}}{6} + x^{4} + \frac{3}{2} x^{2}

Add a constant to the solution

= \frac{x ^{6}}{6} + x^{4} + \frac{3}{2} x^{2} + C

\frac{d}{d x} (8 e^{2})

n = 1 \sum \infty \frac{1}{n ^{\frac{3}{2}}}

\frac{d}{d x} (\frac{87 x}{99 x - 87})

\frac{s + 1}{s ^{2} - 5 s + 6}

y = x^{3} - 6 x^{2} + 9 x, y = x

What is the integral of (x^2+1)(x^3+3x) ?
The integral of (x^2+1)(x^3+3x) is (x^6)/6+x^4+3/2 x^2+C