What is the integral of x(2x^2+1)^6 ?

The integral of x(2x^2+1)^6 is 1/28 (2x^2+1)^7+C

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integral of x(2x^2+1)^6

\int x (2 x^{2} + 1)^{6} d x

\frac{1}{28} (2 x^{2} + 1)^{7} + C

Solution steps

\int x (2 x^{2} + 1)^{6} d x

Apply u-substitution

: \int \frac{( 2 u + 1 ) ^{6}}{2} d u

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

Apply u-substitution

: \int \frac{v ^{6}}{2} d v

= \frac{1}{2} \cdot \int \frac{v ^{6}}{2} d v

Take the constant out:

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

= \frac{1}{2} \cdot \frac{1}{2} \cdot \int v^{6} d v

Apply the Power Rule

: \frac{v ^{7}}{7}

= \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{v ^{7}}{7}

Substitute back

= \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{( 2 x ^{2} + 1 ) ^{7}}{7}

Simplify

\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{( 2 x ^{2} + 1 ) ^{7}}{7} : \frac{1}{28} (2 x^{2} + 1)^{7}

= \frac{1}{28} (2 x^{2} + 1)^{7}

Add a constant to the solution

= \frac{1}{28} (2 x^{2} + 1)^{7} + C

t + 1

\frac{d}{d x} (\frac{2 x}{x + 9})

\frac{\partial}{\partial x} (8 - x^{2} + x y - 4 y^{2})

\int \frac{ln ( x )}{e ^{x}} d x

What is the integral of x(2x^2+1)^6 ?
The integral of x(2x^2+1)^6 is 1/28 (2x^2+1)^7+C