What is the integral of (x^3)/((x^4+3)^{(7/2))} ?

The integral of (x^3)/((x^4+3)^{(7/2))} is -1/(10(x^4+3)^{5/2)}+C

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integral of (x^3)/((x^4+3)^{(7/2))}

\int \frac{x ^{3}}{( x ^{4} + 3 ) ^{(\frac{7}{2})}} d x

- \frac{1}{10 ( x ^{4} + 3 ) ^{\frac{5}{2}}} + C

Solution steps

\int \frac{x ^{3}}{( x ^{4} + 3 ) ^{\frac{7}{2}}} d x

Apply u-substitution

: \int \frac{1}{4 u ^{\frac{7}{2}}} d u

= \int \frac{1}{4 u ^{\frac{7}{2}}} 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{1}{u ^{\frac{7}{2}}} d u

Apply the Power Rule

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

= \frac{1}{4} (- \frac{2}{5 u ^{\frac{5}{2}}})

Substitute back

u = x^{4} + 3

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

Simplify

\frac{1}{4} (- \frac{2}{5 ( x ^{4} + 3 ) ^{\frac{5}{2}}}) : - \frac{1}{10 ( x ^{4} + 3 ) ^{\frac{5}{2}}}

= - \frac{1}{10 ( x ^{4} + 3 ) ^{\frac{5}{2}}}

Add a constant to the solution

= - \frac{1}{10 ( x ^{4} + 3 ) ^{\frac{5}{2}}} + C

\int \frac{1}{( 2 + x ) ^{2}} d x

\int_{0}^{\infty} e^{- a x} d x

\frac{d}{d x} (ln (2 x^{2} y))

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

\frac{d}{d x} (\frac{x - 1}{x ^{2} - 3 x + 3})

What is the integral of (x^3)/((x^4+3)^{(7/2))} ?
The integral of (x^3)/((x^4+3)^{(7/2))} is -1/(10(x^4+3)^{5/2)}+C