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

The integral of (3x^2+1)^3 is 1/35 x(135x^6+189x^4+105x^2+35)+C

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

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

\frac{1}{35} x (135 x^{6} + 189 x^{4} + 105 x^{2} + 35) + C

Solution steps

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

Apply Integral Substitution

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

Take the constant out:

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

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

Expand

(u^{2} + 1)^{3} : u^{6} + 3 u^{4} + 3 u^{2} + 1

= \frac{1}{3} \cdot \int u^{6} + 3 u^{4} + 3 u^{2} + 1 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}{3} (\int u^{6} d u + \int 3 u^{4} d u + \int 3 u^{2} d u + \int 1 d u)

\int u^{6} d u = \frac{u ^{7}}{7}

\int 3 u^{4} d u = \frac{3 u ^{5}}{5}

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

\int 1 d u = u

= \frac{1}{3} (\frac{u ^{7}}{7} + \frac{3 u ^{5}}{5} + u^{3} + u)

Substitute back

u = 3 x

= \frac{1}{3} (\frac{( 3 x ) ^{7}}{7} + \frac{3 ( 3 x ) ^{5}}{5} + (3 x)^{3} + 3 x)

Simplify

\frac{1}{3} (\frac{( 3 x ) ^{7}}{7} + \frac{3 ( 3 x ) ^{5}}{5} + (3 x)^{3} + 3 x) : \frac{1}{35} x (135 x^{6} + 189 x^{4} + 105 x^{2} + 35)

= \frac{1}{35} x (135 x^{6} + 189 x^{4} + 105 x^{2} + 35)

Add a constant to the solution

= \frac{1}{35} x (135 x^{6} + 189 x^{4} + 105 x^{2} + 35) + C

\frac{d y}{d x} + y cos (x) = 2 cos (x), y (0) = 4

d er i v a t i v e sec (θ)

\int 4 x ln (2 x) d x

\int sin^{3} (θ) cos^{4} (θ) d θ

\frac{d y}{d x} = \frac{y + x ^{2} + y ^{2}}{x}

What is the integral of (3x^2+1)^3 ?
The integral of (3x^2+1)^3 is 1/35 x(135x^6+189x^4+105x^2+35)+C