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

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

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

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

\frac{1}{24} (3 + x^{4})^{6} + C

Solution steps

\int x^{3} (3 + 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 = 3 + x^{4}

= \frac{1}{4} \cdot \frac{( 3 + x ^{4} ) ^{6}}{6}

Simplify

\frac{1}{4} \cdot \frac{( 3 + x ^{4} ) ^{6}}{6} : \frac{1}{24} (3 + x^{4})^{6}

= \frac{1}{24} (3 + x^{4})^{6}

Add a constant to the solution

= \frac{1}{24} (3 + x^{4})^{6} + C

\frac{\partial}{\partial x} (2 x^{2} y + 3)

\frac{x + 1}{x - 1}

\frac{d r}{d θ} r = (4 + sec (θ)) sin (θ)

f (x) = (x^{2} \cdot 5^{x})^{9}

\int 12 x (3 x^{2} - 1)^{3} d x

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