What is the integral of (6400x)/(16x^2+3200) ?

The integral of (6400x)/(16x^2+3200) is 200ln|16x^2+3200|+C

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integral of (6400x)/(16x^2+3200)

\int \frac{6400 x}{16 x ^{2} + 3200} d x

200 ln 16 x^{2} + 3200 + C

Solution steps

\int \frac{6400 x}{16 x ^{2} + 3200} d x

Take the constant out:

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

= 6400 \cdot \int \frac{x}{16 x ^{2} + 3200} d x

Apply u-substitution

: \int \frac{1}{32 u} d u

= 6400 \cdot \int \frac{1}{32 u} d u

Take the constant out:

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

= 6400 \cdot \frac{1}{32} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 6400 \cdot \frac{1}{32} ln ∣ u ∣

Substitute back

u = 16 x^{2} + 3200

= 6400 \cdot \frac{1}{32} ln 16 x^{2} + 3200

Simplify

6400 \cdot \frac{1}{32} ln 16 x^{2} + 3200 : 200 ln 16 x^{2} + 3200

= 200 ln 16 x^{2} + 3200

Add a constant to the solution

= 200 ln 16 x^{2} + 3200 + C

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What is the integral of (6400x)/(16x^2+3200) ?
The integral of (6400x)/(16x^2+3200) is 200ln|16x^2+3200|+C