What is the integral of (10)/(sqrt(x^2+16)) ?

The integral of (10)/(sqrt(x^2+16)) is 10ln(1/4 |x+sqrt(16+x^2)|)+C

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integral of (10)/(sqrt(x^2+16))

\int \frac{10}{x ^{2} + 16} d x

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

Solution steps

\int \frac{10}{x ^{2} + 16} d x

Take the constant out:

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

= 10 \cdot \int \frac{1}{x ^{2} + 16} d x

Apply Trigonometric Substitution

: \int sec (u) d u

= 10 \cdot \int sec (u) d u

Use the common integral:

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

= 10 ln ∣ tan (u) + sec (u) ∣

Substitute back

u = arctan (\frac{1}{4} x)

= 10 ln tan (arctan (\frac{1}{4} x)) + sec (arctan (\frac{1}{4} x))

Simplify

10 ln tan (arctan (\frac{1}{4} x)) + sec (arctan (\frac{1}{4} x)) : 10 ln (\frac{1}{4} x + 16 + x^{2})

= 10 ln (\frac{1}{4} x + 16 + x^{2})

Add a constant to the solution

= 10 ln (\frac{1}{4} x + 16 + x^{2}) + C

\frac{d}{d x} (\frac{1}{15} cos^{3} (x) (3 cos^{2} (x) - 5))

\int_{4}^{e^{5}} \frac{1}{x} d x

\frac{d ^{3}}{d x ^{3}} (\frac{1}{1 + x})

y = \frac{4 x}{e ^{x}}

\int_{- 1}^{x^{3}} t^{5} d t

What is the integral of (10)/(sqrt(x^2+16)) ?
The integral of (10)/(sqrt(x^2+16)) is 10ln(1/4 |x+sqrt(16+x^2)|)+C