What is the integral of (200)/(1+x) ?

The integral of (200)/(1+x) is 200ln|1+x|+C

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integral of (200)/(1+x)

\int \frac{200}{1 + x} d x

200 ln ∣ 1 + x ∣ + C

Solution steps

\int \frac{200}{1 + x} d x

Take the constant out:

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

= 200 \cdot \int \frac{1}{1 + x} d x

Apply u-substitution

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

= 200 \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 200 ln ∣ u ∣

Substitute back

u = 1 + x

= 200 ln ∣ 1 + x ∣

Add a constant to the solution

= 200 ln ∣ 1 + x ∣ + C

\frac{1 + 5 s ^{2}}{s ^{2} ( s - 9 )}

\frac{3}{2 s}

\int x^{2} - x + 1 d x

\int 8 e^{5 x} d x

- x^{2} + 6 x, x^{2} - 4 x

What is the integral of (200)/(1+x) ?
The integral of (200)/(1+x) is 200ln|1+x|+C