What is the integral of (x^{n-1})/(sqrt(a+bx^n)) ?

The integral of (x^{n-1})/(sqrt(a+bx^n)) is (2sqrt(a+bx^n))/(bn)+C

Popular >

integral of (x^{n-1})/(sqrt(a+bx^n))

\int \frac{x ^{n - 1}}{a + b x ^{n}} d x

\frac{2 a + b x ^{n}}{bn} + C

Solution steps

\int \frac{x ^{n - 1}}{a + b x ^{n}} d x

Apply u-substitution

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

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

Take the constant out:

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

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

Apply the Power Rule

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

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

Substitute back

u = a + b x^{n}

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

Simplify

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

= \frac{2 a + b x ^{n}}{bn}

Add a constant to the solution

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

\int_{0}^{1} \frac{e ^{- x}}{x} d x

\frac{d y}{d x} = \frac{y ^{2} x}{x ^{2} + 1}

6 x - 2 x^{2} = y, (3, 4)

y^{^{''}} - 8 y^{^{'}} + 16 y = \frac{12 e ^{4 t}}{( t ^{2} + 1 )}

y = sin (x) cos (x), at x = \frac{3 π}{4}

What is the integral of (x^{n-1})/(sqrt(a+bx^n)) ?
The integral of (x^{n-1})/(sqrt(a+bx^n)) is (2sqrt(a+bx^n))/(bn)+C