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

The integral of (x+3)/((3x-4)^{3/2)} is (6x-34)/(9(3x-4)^{1/2)}+C

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

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

\frac{6 x - 34}{9 ( 3 x - 4 ) ^{\frac{1}{2}}} + C

Solution steps

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

Apply u-substitution

: \int \frac{u + 13}{9 u ^{\frac{3}{2}}} d u

= \int \frac{u + 13}{9 u ^{\frac{3}{2}}} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int \frac{u + 13}{u ^{\frac{3}{2}}} d u

Expand

\frac{u + 13}{u ^{\frac{3}{2}}} : \frac{1}{u ^{\frac{1}{2}}} + \frac{13}{u ^{\frac{3}{2}}}

= \frac{1}{9} \cdot \int \frac{1}{u ^{\frac{1}{2}}} + \frac{13}{u ^{\frac{3}{2}}} 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}{9} (\int \frac{1}{u ^{\frac{1}{2}}} d u + \int \frac{13}{u ^{\frac{3}{2}}} d u)

\int \frac{1}{u ^{\frac{1}{2}}} d u = 2 u^{\frac{1}{2}}

\int \frac{13}{u ^{\frac{3}{2}}} d u = - \frac{26}{u ^{\frac{1}{2}}}

= \frac{1}{9} (2 u^{\frac{1}{2}} - \frac{26}{u ^{\frac{1}{2}}})

Substitute back

u = 3 x - 4

= \frac{1}{9} (2 (3 x - 4)^{\frac{1}{2}} - \frac{26}{( 3 x - 4 ) ^{\frac{1}{2}}})

Simplify

\frac{1}{9} (2 (3 x - 4)^{\frac{1}{2}} - \frac{26}{( 3 x - 4 ) ^{\frac{1}{2}}}) : \frac{6 x - 34}{9 ( 3 x - 4 ) ^{\frac{1}{2}}}

= \frac{6 x - 34}{9 ( 3 x - 4 ) ^{\frac{1}{2}}}

Add a constant to the solution

= \frac{6 x - 34}{9 ( 3 x - 4 ) ^{\frac{1}{2}}} + C

\frac{\partial}{\partial x} (7 x^{4} y - 7 x y^{2})

f (x) = x^{2} - 2, at x = 1

\frac{d}{d x} (6 x^{3} e^{x})

\int_{0}^{\frac{π}{4}} 2 sec^{2} (t) d t

f (x) = \frac{3 x}{1 - x}

What is the integral of (x+3)/((3x-4)^{3/2)} ?
The integral of (x+3)/((3x-4)^{3/2)} is (6x-34)/(9(3x-4)^{1/2)}+C