What is the integral of (4x^6+3x^4)/(x^3) ?

The integral of (4x^6+3x^4)/(x^3) is 1/16 (4x^2+3)^2+C

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integral of (4x^6+3x^4)/(x^3)

\int \frac{4 x ^{6} + 3 x ^{4}}{x ^{3}} d x

\frac{1}{16} (4 x^{2} + 3)^{2} + C

Solution steps

\int \frac{4 x ^{6} + 3 x ^{4}}{x ^{3}} d x

Simplify

\frac{4 x ^{6} + 3 x ^{4}}{x ^{3}} : x (4 x^{2} + 3)

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

Apply u-substitution

: \int \frac{u}{8} d u

= \int \frac{u}{8} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int u d u

Apply the Power Rule

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

= \frac{1}{8} \cdot \frac{u ^{2}}{2}

Substitute back

u = 4 x^{2} + 3

= \frac{1}{8} \cdot \frac{( 4 x ^{2} + 3 ) ^{2}}{2}

Simplify

\frac{1}{8} \cdot \frac{( 4 x ^{2} + 3 ) ^{2}}{2} : \frac{1}{16} (4 x^{2} + 3)^{2}

= \frac{1}{16} (4 x^{2} + 3)^{2}

Add a constant to the solution

= \frac{1}{16} (4 x^{2} + 3)^{2} + C

y^{'} = 0.0018 y (2400 - y)

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

ln ∣ ln (x) ∣

\frac{d y}{d t} + \frac{1}{20} y = \frac{3}{2}, y (0) = 0

\frac{d}{d t} (t cos (t) - sin (t))

What is the integral of (4x^6+3x^4)/(x^3) ?
The integral of (4x^6+3x^4)/(x^3) is 1/16 (4x^2+3)^2+C