What is the integral of (x+4)/(sqrt(x+1)) ?

The integral of (x+4)/(sqrt(x+1)) is 2(1/3 (x+1)^{3/2}+3sqrt(x+1))+C

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integral of (x+4)/(sqrt(x+1))

\int \frac{x + 4}{x + 1} d x

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

Solution steps

\int \frac{x + 4}{x + 1} d x

Apply u-substitution

: \int 2 (u^{2} + 3) d u

= \int 2 (u^{2} + 3) d u

Take the constant out:

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

= 2 \cdot \int u^{2} + 3 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

= 2 (\int u^{2} d u + \int 3 d u)

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

\int 3 d u = 3 u

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

Substitute back

u = x + 1

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

Simplify

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

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

Add a constant to the solution

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

\int (\frac{1}{x ( 1 + ln ( x ) )}) d x

\frac{d}{d x} (e^{- x} cos (x))

\int_{0}^{2} x x^{2} + 2 d x

\int_{0}^{4} (2 x + 5) d x

x^{2} f (x)

What is the integral of (x+4)/(sqrt(x+1)) ?
The integral of (x+4)/(sqrt(x+1)) is 2(1/3 (x+1)^{3/2}+3sqrt(x+1))+C