What is the integral of x^3(7+x^4)^5 ?

The integral of x^3(7+x^4)^5 is 1/24 (7+x^4)^6+C

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integral of x^3(7+x^4)^5

\int x^{3} (7 + x^{4})^{5} d x

\frac{1}{24} (7 + x^{4})^{6} + C

Solution steps

\int x^{3} (7 + x^{4})^{5} d x

Apply u-substitution

: \int \frac{u ^{5}}{4} d u

= \int \frac{u ^{5}}{4} 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 u^{5} d u

Apply the Power Rule

: \frac{u ^{6}}{6}

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

Substitute back

u = 7 + x^{4}

= \frac{1}{4} \cdot \frac{( 7 + x ^{4} ) ^{6}}{6}

Simplify

\frac{1}{4} \cdot \frac{( 7 + x ^{4} ) ^{6}}{6} : \frac{1}{24} (7 + x^{4})^{6}

= \frac{1}{24} (7 + x^{4})^{6}

Add a constant to the solution

= \frac{1}{24} (7 + x^{4})^{6} + C

\int_{0}^{\frac{π}{3}} ln (sec (x)) d x

\frac{d}{d x} (\frac{e ^{x}}{8 + e ^{x}})

\int - 14 x^{- 4} y^{6} + 16 x^{- 5} y^{7} d y

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

y^{^{''}} - 14 y^{^{'}} + 51 y = 0

What is the integral of x^3(7+x^4)^5 ?
The integral of x^3(7+x^4)^5 is 1/24 (7+x^4)^6+C