What is the integral of (10x)/(sqrt(64x^4-49)) ?

The integral of (10x)/(sqrt(64x^4-49)) is 5/8 ln((|8x^2+sqrt(64x^4-49)|)/7)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (10x)/(sqrt(64x^4-49))

\int \frac{10 x}{64 x ^{4} - 49} d x

\frac{5}{8} ln (\frac{8 x ^{2} + 64 x ^{4} - 49}{7}) + C

Solution steps

\int \frac{10 x}{64 x ^{4} - 49} d x

Take the constant out:

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

= 10 \cdot \int \frac{x}{64 x ^{4} - 49} d x

Apply u-substitution

: \int \frac{1}{2 64 u ^{2} - 49} d u

= 10 \cdot \int \frac{1}{2 64 u ^{2} - 49} d u

Take the constant out:

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

= 10 \cdot \frac{1}{2} \cdot \int \frac{1}{64 u ^{2} - 49} d u

Apply Trigonometric Substitution

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

= 10 \cdot \frac{1}{2} \cdot \int \frac{sec ( v )}{8} d v

Take the constant out:

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

= 10 \cdot \frac{1}{2} \cdot \frac{1}{8} \cdot \int sec (v) d v

Use the common integral:

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

= 10 \cdot \frac{1}{2} \cdot \frac{1}{8} ln ∣ tan (v) + sec (v) ∣

Substitute back

= 10 \cdot \frac{1}{2} \cdot \frac{1}{8} ln tan (arcsec (\frac{8}{7} x^{2})) + sec (arcsec (\frac{8}{7} x^{2}))

Simplify

10 \cdot \frac{1}{2} \cdot \frac{1}{8} ln tan (arcsec (\frac{8}{7} x^{2})) + sec (arcsec (\frac{8}{7} x^{2})) : \frac{5}{8} ln (\frac{8 x ^{2} + 64 x ^{4} - 49}{7})

= \frac{5}{8} ln (\frac{8 x ^{2} + 64 x ^{4} - 49}{7})

Add a constant to the solution

= \frac{5}{8} ln (\frac{8 x ^{2} + 64 x ^{4} - 49}{7}) + C

\frac{d}{d x} (sin (x) tan (x))

((f (x))^{2})^{'}

\frac{d}{d x} (cos (2 x) e^{2 x})

x \to - 1 lim (\frac{2 x + 3}{x + 1})

What is the integral of (10x)/(sqrt(64x^4-49)) ?
The integral of (10x)/(sqrt(64x^4-49)) is 5/8 ln((|8x^2+sqrt(64x^4-49)|)/7)+C