What is the integral of x/((4-9x^4)^{3/2)} ?

The integral of x/((4-9x^4)^{3/2)} is (x^2)/(8(-9x^4+4)^{1/2)}+C

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integral of x/((4-9x^4)^{3/2)}

\int \frac{x}{( 4 - 9 x ^{4} ) ^{\frac{3}{2}}} d x

\frac{x ^{2}}{8 ( - 9 x ^{4} + 4 ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{x}{( 4 - 9 x ^{4} ) ^{\frac{3}{2}}} d x

Apply u-substitution

: \int \frac{1}{2 ( 4 - 9 u ^{2} ) ^{\frac{3}{2}}} d u

= \int \frac{1}{2 ( 4 - 9 u ^{2} ) ^{\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}{2} \cdot \int \frac{1}{( 4 - 9 u ^{2} ) ^{\frac{3}{2}}} d u

Apply Trigonometric Substitution

: \int \frac{1}{12 cos ^{2} ( v )} d v

= \frac{1}{2} \cdot \int \frac{1}{12 cos ^{2} ( v )} 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}{12} \cdot \int \frac{1}{cos ^{2} ( v )} d v

Use the common integral:

\int \frac{1}{cos ^{2} ( v )} d v = tan (v)

= \frac{1}{2} \cdot \frac{1}{12} tan (v)

Substitute back

= \frac{1}{2} \cdot \frac{1}{12} tan (arcsin (\frac{3}{2} x^{2}))

Simplify

\frac{1}{2} \cdot \frac{1}{12} tan (arcsin (\frac{3}{2} x^{2})) : \frac{x ^{2}}{8 ( - 9 x ^{4} + 4 ) ^{\frac{1}{2}}}

= \frac{x ^{2}}{8 ( - 9 x ^{4} + 4 ) ^{\frac{1}{2}}}

Add a constant to the solution

= \frac{x ^{2}}{8 ( - 9 x ^{4} + 4 ) ^{\frac{1}{2}}} + C

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What is the integral of x/((4-9x^4)^{3/2)} ?
The integral of x/((4-9x^4)^{3/2)} is (x^2)/(8(-9x^4+4)^{1/2)}+C