What is the integral of (8x)/(2x^2+3)ln(2x^2+3) ?

The integral of (8x)/(2x^2+3)ln(2x^2+3) is ln^2(3+2x^2)+C

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integral of (8x)/(2x^2+3)ln(2x^2+3)

\int \frac{8 x}{2 x ^{2} + 3} ln (2 x^{2} + 3) d x

ln^{2} (3 + 2 x^{2}) + C

Solution steps

\int \frac{8 x}{2 x ^{2} + 3} ln (2 x^{2} + 3) 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 x ^{2} + 3} ln (2 x^{2} + 3) d x

\frac{x}{2 x ^{2} + 3} = \frac{x - 2 x ^{2} - 3}{2 x ^{2} + 3} + 1

= 8 \cdot \int (\frac{x - 2 x ^{2} - 3}{2 x ^{2} + 3} + 1) ln (2 x^{2} + 3) d x

Apply u-substitution

= 8 \cdot \int \frac{u}{4} d u

Take the constant out:

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

= 8 \cdot \frac{1}{4} \cdot \int u d u

Apply the Power Rule

= 8 \cdot \frac{1}{4} \cdot \frac{u ^{2}}{2}

Substitute back

u = ln (3 + 2 x^{2})

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

Simplify

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

= ln^{2} (3 + 2 x^{2})

Add a constant to the solution

= ln^{2} (3 + 2 x^{2}) + C

What is the integral of (8x)/(2x^2+3)ln(2x^2+3) ?
The integral of (8x)/(2x^2+3)ln(2x^2+3) is ln^2(3+2x^2)+C