What is the integral of x^2In(4x) ?

The integral of x^2In(4x) is Inx^4+C

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integral of x^2In(4x)

\int x^{2} I n (4 x) d x

I n x^{4} + C

Solution steps

\int x^{2} I n \cdot 4 x d x

Take the constant out:

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

= 4 I n \cdot \int x x^{2} d x

Simplify

x x^{2} : x^{3}

= 4 I n \cdot \int x^{3} d x

Apply the Power Rule

: \frac{x ^{4}}{4}

= 4 I n \frac{x ^{4}}{4}

Simplify

4 I n \frac{x ^{4}}{4} : I n x^{4}

= I n x^{4}

Add a constant to the solution

= I n x^{4} + C

f (x) = x^{\frac{4}{5}}

f (x) = arcsin (x) + x^{2}, at x = 0

\int cos^{4} (7 x + 9) d x

\frac{d}{d x} (2 x^{2} e^{2 x})

\frac{x ^{2}}{1 + 8 x}