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

The integral of (3x^2+1)^3x is 1/24 (3x^2+1)^4+C

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

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

\frac{1}{24} (3 x^{2} + 1)^{4} + C

Solution steps

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

Apply u-substitution

: \int \frac{u ^{3}}{6} d u

= \int \frac{u ^{3}}{6} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int u^{3} d u

Apply the Power Rule

: \frac{u ^{4}}{4}

= \frac{1}{6} \cdot \frac{u ^{4}}{4}

Substitute back

u = 3 x^{2} + 1

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

Simplify

\frac{1}{6} \cdot \frac{( 3 x ^{2} + 1 ) ^{4}}{4} : \frac{1}{24} (3 x^{2} + 1)^{4}

= \frac{1}{24} (3 x^{2} + 1)^{4}

Add a constant to the solution

= \frac{1}{24} (3 x^{2} + 1)^{4} + C

x \to - 1 lim (\frac{x}{x ^{2} + 1})

\int_{0}^{6} 2 x + 1 d x

x \to 0 lim (sin (x) \cdot \frac{1}{x})

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

x \to 8 lim (\frac{( x + 6 )}{x - 8})

What is the integral of (3x^2+1)^3x ?
The integral of (3x^2+1)^3x is 1/24 (3x^2+1)^4+C