What is the integral of 6x(6x^2+1)^{5/2} ?

The integral of 6x(6x^2+1)^{5/2} is 1/7 (6x^2+1)^{7/2}+C

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integral of 6x(6x^2+1)^{5/2}

\int 6 x (6 x^{2} + 1)^{\frac{5}{2}} d x

\frac{1}{7} (6 x^{2} + 1)^{\frac{7}{2}} + C

Solution steps

\int 6 x (6 x^{2} + 1)^{\frac{5}{2}} d x

Take the constant out:

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

= 6 \cdot \int x (6 x^{2} + 1)^{\frac{5}{2}} d x

Apply u-substitution

: \int \frac{( 6 u + 1 ) ^{\frac{5}{2}}}{2} d u

= 6 \cdot \int \frac{( 6 u + 1 ) ^{\frac{5}{2}}}{2} d u

Take the constant out:

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

= 6 \cdot \frac{1}{2} \cdot \int (6 u + 1)^{\frac{5}{2}} d u

Apply u-substitution

: \int \frac{v ^{\frac{5}{2}}}{6} d v

= 6 \cdot \frac{1}{2} \cdot \int \frac{v ^{\frac{5}{2}}}{6} d v

Take the constant out:

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

= 6 \cdot \frac{1}{2} \cdot \frac{1}{6} \cdot \int v^{\frac{5}{2}} d v

Apply the Power Rule

: \frac{2}{7} v^{\frac{7}{2}}

= 6 \cdot \frac{1}{2} \cdot \frac{1}{6} \cdot \frac{2}{7} v^{\frac{7}{2}}

Substitute back

= 6 \cdot \frac{1}{2} \cdot \frac{1}{6} \cdot \frac{2}{7} (6 x^{2} + 1)^{\frac{7}{2}}

Simplify

6 \cdot \frac{1}{2} \cdot \frac{1}{6} \cdot \frac{2}{7} (6 x^{2} + 1)^{\frac{7}{2}} : \frac{1}{7} (6 x^{2} + 1)^{\frac{7}{2}}

= \frac{1}{7} (6 x^{2} + 1)^{\frac{7}{2}}

Add a constant to the solution

= \frac{1}{7} (6 x^{2} + 1)^{\frac{7}{2}} + C

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What is the integral of 6x(6x^2+1)^{5/2} ?
The integral of 6x(6x^2+1)^{5/2} is 1/7 (6x^2+1)^{7/2}+C