What is the integral of ((3x^2-4)^{3/2}4x^3)/(x^2) ?

The integral of ((3x^2-4)^{3/2}4x^3)/(x^2) is 4/15 (3x^2-4)^{5/2}+C

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integral of ((3x^2-4)^{3/2}4x^3)/(x^2)

\int \frac{( 3 x ^{2} - 4 ) ^{\frac{3}{2}} 4 x ^{3}}{x ^{2}} d x

\frac{4}{15} (3 x^{2} - 4)^{\frac{5}{2}} + C

Solution steps

\int \frac{( 3 x ^{2} - 4 ) ^{\frac{3}{2}} \cdot 4 x ^{3}}{x ^{2}} d x

Take the constant out:

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

= 4 \cdot \int \frac{( 3 x ^{2} - 4 ) ^{\frac{3}{2}} x ^{3}}{x ^{2}} d x

Simplify

\frac{( 3 x ^{2} - 4 ) ^{\frac{3}{2}} x ^{3}}{x ^{2}} : x (3 x^{2} - 4)^{\frac{3}{2}}

= 4 \cdot \int x (3 x^{2} - 4)^{\frac{3}{2}} d x

Apply u-substitution

: \int \frac{u ^{\frac{3}{2}}}{6} d u

= 4 \cdot \int \frac{u ^{\frac{3}{2}}}{6} d u

Take the constant out:

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

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

Apply the Power Rule

: \frac{2}{5} u^{\frac{5}{2}}

= 4 \cdot \frac{1}{6} \cdot \frac{2}{5} u^{\frac{5}{2}}

Substitute back

u = 3 x^{2} - 4

= 4 \cdot \frac{1}{6} \cdot \frac{2}{5} (3 x^{2} - 4)^{\frac{5}{2}}

Simplify

4 \cdot \frac{1}{6} \cdot \frac{2}{5} (3 x^{2} - 4)^{\frac{5}{2}} : \frac{4}{15} (3 x^{2} - 4)^{\frac{5}{2}}

= \frac{4}{15} (3 x^{2} - 4)^{\frac{5}{2}}

Add a constant to the solution

= \frac{4}{15} (3 x^{2} - 4)^{\frac{5}{2}} + C

\frac{d}{d x} (\frac{sin ( x )}{1 + cos ( x )})

\int e^{- s t} t cos (t) d t

\frac{5}{s ( s ^{2} + s + 1 )}

\int_{- 2}^{2} 4^{x} d x

\frac{( d y )}{( d x )} = 8 x^{(3)} + 6 x^{(2)}

What is the integral of ((3x^2-4)^{3/2}4x^3)/(x^2) ?
The integral of ((3x^2-4)^{3/2}4x^3)/(x^2) is 4/15 (3x^2-4)^{5/2}+C