What is the integral of x^2|8x^3+1| ?

The integral of x^2|8x^3+1| is 1/48 |8x^3+1|(8x^3+1)+C

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integral of x^2|8x^3+1|

\int x^{2} 8 x^{3} + 1 d x

\frac{1}{48} 8 x^{3} + 1 (8 x^{3} + 1) + C

Solution steps

\int x^{2} 8 x^{3} + 1 d x

Apply u-substitution

: \int \frac{u 8 u ^{\frac{3}{2}} + 1}{2 x} d u

= \int \frac{∣ 8 u + 1 ∣}{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 ∣ 8 u + 1 ∣ d u

Apply u-substitution

: \int ∣ v ∣ \frac{1}{8} d v

= \frac{1}{3} \cdot \int ∣ v ∣ \frac{1}{8} d v

Take the constant out:

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

= \frac{1}{3} \cdot \frac{1}{8} \cdot \int ∣ v ∣ d v

Use the common integral:

\int ∣ v ∣ d v = \frac{v ∣ v ∣}{2}

= \frac{1}{3} \cdot \frac{1}{8} \cdot \frac{v ∣ v ∣}{2}

Substitute back

= \frac{1}{3} \cdot \frac{1}{8} \cdot \frac{( 8 x ^{3} + 1 ) 8 x ^{3} + 1}{2}

Simplify

\frac{1}{3} \cdot \frac{1}{8} \cdot \frac{( 8 x ^{3} + 1 ) 8 x ^{3} + 1}{2} : \frac{1}{48} 8 x^{3} + 1 (8 x^{3} + 1)

= \frac{1}{48} 8 x^{3} + 1 (8 x^{3} + 1)

Add a constant to the solution

= \frac{1}{48} 8 x^{3} + 1 (8 x^{3} + 1) + C

\frac{d}{d x} (2 sec (x) - 7 x)

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\frac{d}{d x} (32)

What is the integral of x^2|8x^3+1| ?
The integral of x^2|8x^3+1| is 1/48 |8x^3+1|(8x^3+1)+C