What is the integral of (4x^2+1/((4x)^2)) ?

The integral of (4x^2+1/((4x)^2)) is (4x^3)/3+1/(16x)-1/(8x)+C

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integral of (4x^2+1/((4x)^2))

\int (4 x^{2} + \frac{1}{( 4 x ) ^{2}}) d x

\frac{4 x ^{3}}{3} + \frac{1}{16 x} - \frac{1}{8 x} + C

Solution steps

\int 4 x^{2} + \frac{1}{( 4 x ) ^{2}} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 4 x^{2} d x + \int \frac{1}{( 4 x ) ^{2}} d x

\int 4 x^{2} d x = \frac{4 x ^{3}}{3}

\int \frac{1}{( 4 x ) ^{2}} d x = \frac{1}{16 x} - \frac{1}{8 x}

= \frac{4 x ^{3}}{3} + \frac{1}{16 x} - \frac{1}{8 x}

Add a constant to the solution

= \frac{4 x ^{3}}{3} + \frac{1}{16 x} - \frac{1}{8 x} + C

y^{^{'}} = 9 x 7 y + 8, y (0) = 3

\frac{d}{d x} (\frac{x \cdot cos ( x ) + π}{π - x})

\frac{t}{3}

\int (sin^{4} (3 x) - cos^{4} (3 x)) d x

\int r^{3} 1 + 4 r^{2} d r

What is the integral of (4x^2+1/((4x)^2)) ?
The integral of (4x^2+1/((4x)^2)) is (4x^3)/3+1/(16x)-1/(8x)+C