What is the integral of (6x^2)/(sqrt(4x^6+4)) ?

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

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integral of (6x^2)/(sqrt(4x^6+4))

\int \frac{6 x ^{2}}{4 x ^{6} + 4} d x

ln x^{6} + 1 + x^{3} + C

Solution steps

\int \frac{6 x ^{2}}{4 x ^{6} + 4} d x

Take the constant out:

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

= 6 \cdot \int \frac{x ^{2}}{4 x ^{6} + 4} d x

4 x^{6} + 4 = 2 x^{6} + 1

= 6 \cdot \int \frac{x ^{2}}{2 x ^{6} + 1} d x

Take the constant out:

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

= 6 \cdot \frac{1}{2} \cdot \int \frac{x ^{2}}{x ^{6} + 1} d x

Apply u-substitution

: \int \frac{1}{3 u ^{2} + 1} d u

= 6 \cdot \frac{1}{2} \cdot \int \frac{1}{3 u ^{2} + 1} d u

Take the constant out:

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

= 6 \cdot \frac{1}{2} \cdot \frac{1}{3} \cdot \int \frac{1}{u ^{2} + 1} d u

Use the common integral:

\int \frac{1}{u ^{2} + 1} d u = ln u^{2} + 1 + u

= 6 \cdot \frac{1}{2} \cdot \frac{1}{3} ln u^{2} + 1 + u

Substitute back

u = x^{3}

= 6 \cdot \frac{1}{2} \cdot \frac{1}{3} ln (x^{3})^{2} + 1 + x^{3}

Simplify

6 \cdot \frac{1}{2} \cdot \frac{1}{3} ln (x^{3})^{2} + 1 + x^{3} : ln x^{6} + 1 + x^{3}

= ln x^{6} + 1 + x^{3}

Add a constant to the solution

= ln x^{6} + 1 + x^{3} + C

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