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

The integral of (7x^3)/(sqrt(4x+5)) is 1/8 (4x^3-6x^2+10x-25)sqrt(4x+5)+C

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

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

\frac{1}{8} (4 x^{3} - 6 x^{2} + 10 x - 25) 4 x + 5 + C

Solution steps

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

Apply u-substitution

: \int \frac{( u ^{2} - 5 ) ^{3}}{128} d u

= 7 \cdot \int \frac{( u ^{2} - 5 ) ^{3}}{128} d u

Take the constant out:

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

= 7 \cdot \frac{1}{128} \cdot \int (u^{2} - 5)^{3} d u

Expand

(u^{2} - 5)^{3} : u^{6} - 15 u^{4} + 75 u^{2} - 125

= 7 \cdot \frac{1}{128} \cdot \int u^{6} - 15 u^{4} + 75 u^{2} - 125 d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 7 \cdot \frac{1}{128} (\int u^{6} d u - \int 15 u^{4} d u + \int 75 u^{2} d u - \int 125 d u)

\int u^{6} d u = \frac{u ^{7}}{7}

\int 15 u^{4} d u = 3 u^{5}

\int 75 u^{2} d u = 25 u^{3}

\int 125 d u = 125 u

= 7 \cdot \frac{1}{128} (\frac{u ^{7}}{7} - 3 u^{5} + 25 u^{3} - 125 u)

Substitute back

u = 4 x + 5

= 7 \cdot \frac{1}{128} (\frac{( 4 x + 5 ) ^{7}}{7} - 3 (4 x + 5)^{5} + 25 (4 x + 5)^{3} - 125 4 x + 5)

Simplify

7 \cdot \frac{1}{128} (\frac{( 4 x + 5 ) ^{7}}{7} - 3 (4 x + 5)^{5} + 25 (4 x + 5)^{3} - 125 4 x + 5) : \frac{1}{8} (4 x^{3} - 6 x^{2} + 10 x - 25) 4 x + 5

= \frac{1}{8} (4 x^{3} - 6 x^{2} + 10 x - 25) 4 x + 5

Add a constant to the solution

= \frac{1}{8} (4 x^{3} - 6 x^{2} + 10 x - 25) 4 x + 5 + C

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What is the integral of (7x^3)/(sqrt(4x+5)) ?
The integral of (7x^3)/(sqrt(4x+5)) is 1/8 (4x^3-6x^2+10x-25)sqrt(4x+5)+C