What is the integral of (7x^2)/(sqrt(4x^3-5)) ?

The integral of (7x^2)/(sqrt(4x^3-5)) is 7/6 sqrt(4x^3-5)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (7x^2)/(sqrt(4x^3-5))

\int \frac{7 x ^{2}}{4 x ^{3} - 5} d x

\frac{7}{6} 4 x^{3} - 5 + C

Solution steps

\int \frac{7 x ^{2}}{4 x ^{3} - 5} d x

Take the constant out:

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

= 7 \cdot \int \frac{x ^{2}}{4 x ^{3} - 5} d x

Apply u-substitution

: \int \frac{1}{12 u} d u

= 7 \cdot \int \frac{1}{12 u} d u

Take the constant out:

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

= 7 \cdot \frac{1}{12} \cdot \int \frac{1}{u} d u

Apply the Power Rule

: 2 u^{\frac{1}{2}}

= 7 \cdot \frac{1}{12} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = 4 x^{3} - 5

= 7 \cdot \frac{1}{12} \cdot 2 (4 x^{3} - 5)^{\frac{1}{2}}

Simplify

7 \cdot \frac{1}{12} \cdot 2 (4 x^{3} - 5)^{\frac{1}{2}} : \frac{7}{6} 4 x^{3} - 5

= \frac{7}{6} 4 x^{3} - 5

Add a constant to the solution

= \frac{7}{6} 4 x^{3} - 5 + C

\int \frac{x ^{2}}{x ^{6} + 36} d x

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

\int 8 cos^{3} (5 x) d x

\int cot (2 x) (cot (2 x) + 3 csc (2 x)) d x

\int (1 - x)^{\frac{- 1}{2}} d x

What is the integral of (7x^2)/(sqrt(4x^3-5)) ?
The integral of (7x^2)/(sqrt(4x^3-5)) is 7/6 sqrt(4x^3-5)+C