What is the integral of (8x^2)/(4x^2+7) ?

The integral of (8x^2)/(4x^2+7) is -sqrt(7)arctan(2/(sqrt(7))x)+2x+C

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integral of (8x^2)/(4x^2+7)

\int \frac{8 x ^{2}}{4 x ^{2} + 7} d x

- 7 arctan (\frac{2}{7} x) + 2 x + C

Solution steps

\int \frac{8 x ^{2}}{4 x ^{2} + 7} d x

Take the constant out:

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

= 8 \cdot \int \frac{x ^{2}}{4 x ^{2} + 7} d x

4 x^{2} + 7 = 7 (\frac{4 x ^{2}}{7} + 1)

= 8 \cdot \int \frac{x ^{2}}{7 ( \frac{4 x ^{2}}{7} + 1 )} d x

Take the constant out:

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

= 8 \cdot \frac{1}{7} \cdot \int \frac{x ^{2}}{\frac{4 x ^{2}}{7} + 1} d x

Apply Integral Substitution

= 8 \cdot \frac{1}{7} \cdot \int \frac{7 7 u ^{2}}{8 ( u ^{2} + 1 )} d u

Take the constant out:

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

= 8 \cdot \frac{1}{7} \cdot \frac{7 7}{8} \cdot \int \frac{u ^{2}}{u ^{2} + 1} d u

\frac{u ^{2}}{u ^{2} + 1} = - \frac{1}{u ^{2} + 1} + 1

= 8 \cdot \frac{1}{7} \cdot \frac{7 7}{8} \cdot \int - \frac{1}{u ^{2} + 1} + 1 d u

Apply the Sum Rule:

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

= 8 \cdot \frac{1}{7} \cdot \frac{7 7}{8} (- \int \frac{1}{u ^{2} + 1} d u + \int 1 d u)

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

\int 1 d u = u

= 8 \cdot \frac{1}{7} \cdot \frac{7 7}{8} (- arctan (u) + u)

Substitute back

u = \frac{2}{7} x

= 8 \cdot \frac{1}{7} \cdot \frac{7 7}{8} (- arctan (\frac{2}{7} x) + \frac{2}{7} x)

Simplify

8 \cdot \frac{1}{7} \cdot \frac{7 7}{8} (- arctan (\frac{2}{7} x) + \frac{2}{7} x) : - 7 arctan (\frac{2}{7} x) + 2 x

= - 7 arctan (\frac{2}{7} x) + 2 x

Add a constant to the solution

= - 7 arctan (\frac{2}{7} x) + 2 x + C

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

e^{s^{2}}

\int x^{3} (x^{4} + 2)^{5} d x

y = x^{5} 3^{x}

- \frac{1}{x ^{2}} + sec^{2} (x)

What is the integral of (8x^2)/(4x^2+7) ?
The integral of (8x^2)/(4x^2+7) is -sqrt(7)arctan(2/(sqrt(7))x)+2x+C