What is the integral of (200)/(sqrt(16x^2+256)) ?

The integral of (200)/(sqrt(16x^2+256)) is 50ln(1/4 |x+sqrt(x^2+16)|)+C

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integral of (200)/(sqrt(16x^2+256))

\int \frac{200}{16 x ^{2} + 256} d x

50 ln (\frac{1}{4} x + x^{2} + 16) + C

Solution steps

\int \frac{200}{16 x ^{2} + 256} 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}{16 x ^{2} + 256} d x

Apply Trigonometric Substitution

= 200 \cdot \int \frac{sec ( u )}{4} d u

Take the constant out:

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

= 200 \cdot \frac{1}{4} \cdot \int sec (u) d u

Use the common integral:

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

= 200 \cdot \frac{1}{4} ln ∣ tan (u) + sec (u) ∣

Substitute back

u = arctan (\frac{4}{16} x)

= 200 \cdot \frac{1}{4} ln tan (arctan (\frac{4}{16} x)) + sec (arctan (\frac{4}{16} x))

Simplify

200 \cdot \frac{1}{4} ln tan (arctan (\frac{4}{16} x)) + sec (arctan (\frac{4}{16} x)) : 50 ln (\frac{1}{4} x + x^{2} + 16)

= 50 ln (\frac{1}{4} x + x^{2} + 16)

Add a constant to the solution

= 50 ln (\frac{1}{4} x + x^{2} + 16) + C

x \to 3 - lim (\frac{1}{3 - x})

\int k e^{3 x} d x

\frac{d}{d x} (x + 5 sin (x))

d er i v a t i v e f (x) = 19 e^{x^{2}}

f (x) = e^{e^{x}}

What is the integral of (200)/(sqrt(16x^2+256)) ?
The integral of (200)/(sqrt(16x^2+256)) is 50ln(1/4 |x+sqrt(x^2+16)|)+C