What is the integral of 1/(sqrt(64x^2+1)) ?

The integral of 1/(sqrt(64x^2+1)) is 1/8 ln|8x+sqrt(1+64x^2)|+C

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integral of 1/(sqrt(64x^2+1))

\int \frac{1}{64 x ^{2} + 1} d x

\frac{1}{8} ln 8 x + 1 + 64 x^{2} + C

Solution steps

\int \frac{1}{64 x ^{2} + 1} d x

Apply Trigonometric Substitution

: \int \frac{sec ( u )}{8} d u

= \int \frac{sec ( u )}{8} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int sec (u) d u

Use the common integral:

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

= \frac{1}{8} ln ∣ tan (u) + sec (u) ∣

Substitute back

u = arctan (8 x)

= \frac{1}{8} ln ∣ tan (arctan (8 x)) + sec (arctan (8 x)) ∣

Simplify

\frac{1}{8} ln ∣ tan (arctan (8 x)) + sec (arctan (8 x)) ∣ : \frac{1}{8} ln 8 x + 1 + 64 x^{2}

= \frac{1}{8} ln 8 x + 1 + 64 x^{2}

Add a constant to the solution

= \frac{1}{8} ln 8 x + 1 + 64 x^{2} + C

6 x^{2} (\frac{d y}{d x}) = (\frac{d y}{d x}) + 9 x e^{- y}

\int 8 x^{- 3} d x

\int_{0}^{2} e^{- x} d x

\int 3 + x^{2} d x

\int (3 x^{2} - 9 x + 2) d x

What is the integral of 1/(sqrt(64x^2+1)) ?
The integral of 1/(sqrt(64x^2+1)) is 1/8 ln|8x+sqrt(1+64x^2)|+C