What is the integral of (x^4)/((4-x^2)^{7/2)} ?

The integral of (x^4)/((4-x^2)^{7/2)} is (x^5)/(20(4-x^2)^{5/2)}+C

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integral of (x^4)/((4-x^2)^{7/2)}

\int \frac{x ^{4}}{( 4 - x ^{2} ) ^{\frac{7}{2}}} d x

\frac{x ^{5}}{20 ( 4 - x ^{2} ) ^{\frac{5}{2}}} + C

Solution steps

\int \frac{x ^{4}}{( 4 - x ^{2} ) ^{\frac{7}{2}}} d x

Apply Trigonometric Substitution

: \int \frac{sin ^{4} ( u )}{4 cos ^{6} ( u )} d u

= \int \frac{sin ^{4} ( u )}{4 cos ^{6} ( u )} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{sin ^{4} ( u )}{cos ^{6} ( u )} d u

Rewrite using trig identities

= \frac{1}{4} \cdot \int \frac{tan ^{4} ( u )}{cos ^{2} ( u )} d u

Use the basic trigonometric identity:

\frac{1}{cos ( x )} = sec (x)

= \frac{1}{4} \cdot \int tan^{4} (u) sec^{2} (u) d u

Rewrite using trig identities

= \frac{1}{4} \cdot \int (1 + tan^{2} (u)) tan^{4} (u) d u

Apply u-substitution

: \int v^{4} d v

= \frac{1}{4} \cdot \int v^{4} d v

Apply the Power Rule

: \frac{v ^{5}}{5}

= \frac{1}{4} \cdot \frac{v ^{5}}{5}

Substitute back

= \frac{1}{4} \cdot \frac{tan ^{5} ( arcsin ( \frac{1}{2} x ) )}{5}

Simplify

\frac{1}{4} \cdot \frac{tan ^{5} ( arcsin ( \frac{1}{2} x ) )}{5} : \frac{x ^{5}}{20 ( 4 - x ^{2} ) ^{\frac{5}{2}}}

= \frac{x ^{5}}{20 ( 4 - x ^{2} ) ^{\frac{5}{2}}}

Add a constant to the solution

= \frac{x ^{5}}{20 ( 4 - x ^{2} ) ^{\frac{5}{2}}} + C

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What is the integral of (x^4)/((4-x^2)^{7/2)} ?
The integral of (x^4)/((4-x^2)^{7/2)} is (x^5)/(20(4-x^2)^{5/2)}+C