What is the integral of (13-13x)/(1-sqrt(x)) ?

The integral of (13-13x)/(1-sqrt(x)) is 26(1/3 x^{3/2}+x/2)+C

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integral of (13-13x)/(1-sqrt(x))

\int \frac{13 - 13 x}{1 - x} d x

26 (\frac{1}{3} x^{\frac{3}{2}} + \frac{x}{2}) + C

Solution steps

\int \frac{13 - 13 x}{1 - x} d x

Apply u-substitution

: \int 26 u (u + 1) d u

= \int 26 u (u + 1) d u

Take the constant out:

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

= 26 \cdot \int u (u + 1) d u

Expand

u (u + 1) : u^{2} + u

= 26 \cdot \int u^{2} + u 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

= 26 (\int u^{2} d u + \int u d u)

\int u^{2} d u = \frac{u ^{3}}{3}

\int u d u = \frac{u ^{2}}{2}

= 26 (\frac{u ^{3}}{3} + \frac{u ^{2}}{2})

Substitute back

u = x

= 26 (\frac{( x ) ^{3}}{3} + \frac{( x ) ^{2}}{2})

Simplify

26 (\frac{( x ) ^{3}}{3} + \frac{( x ) ^{2}}{2}) : 26 (\frac{1}{3} x^{\frac{3}{2}} + \frac{x}{2})

= 26 (\frac{1}{3} x^{\frac{3}{2}} + \frac{x}{2})

Add a constant to the solution

= 26 (\frac{1}{3} x^{\frac{3}{2}} + \frac{x}{2}) + C

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

2 4 - y

\int \frac{1}{x ^{2} - a ^{2}} d x

cos (x + y)

\int \frac{t}{t - 1} d t

What is the integral of (13-13x)/(1-sqrt(x)) ?
The integral of (13-13x)/(1-sqrt(x)) is 26(1/3 x^{3/2}+x/2)+C