What is the integral of ((a+bx^4))/(sqrt(5ax+bx^5)) ?

The integral of ((a+bx^4))/(sqrt(5ax+bx^5)) is 2/5 sqrt(5ax+bx^5)+C

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integral of ((a+bx^4))/(sqrt(5ax+bx^5))

\int \frac{( a + b x ^{4} )}{5 a x + b x ^{5}} d x

\frac{2}{5} 5 a x + b x^{5} + C

Solution steps

\int \frac{a + b x ^{4}}{5 a x + b x ^{5}} d x

Apply u-substitution

: \int \frac{1}{5 u} d u

= \int \frac{1}{5 u} d u

Take the constant out:

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

= \frac{1}{5} \cdot \int \frac{1}{u} d u

Apply the Power Rule

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

= \frac{1}{5} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = 5 a x + b x^{5}

= \frac{1}{5} \cdot 2 (5 a x + b x^{5})^{\frac{1}{2}}

Simplify

\frac{1}{5} \cdot 2 (5 a x + b x^{5})^{\frac{1}{2}} : \frac{2}{5} 5 a x + b x^{5}

= \frac{2}{5} 5 a x + b x^{5}

Add a constant to the solution

= \frac{2}{5} 5 a x + b x^{5} + C

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

x \to 4 lim (g (x))

\int \frac{x}{x - x} d x

\frac{d}{d x} (e^{y^{2} - x^{2}})

y^{^{''}} + y^{^{'}} + 4 y = 0, y (0) = 0, y^{^{'}} (0) = 0

What is the integral of ((a+bx^4))/(sqrt(5ax+bx^5)) ?
The integral of ((a+bx^4))/(sqrt(5ax+bx^5)) is 2/5 sqrt(5ax+bx^5)+C