What is the integral of 1/(sqrt(x+4)+\sqrt{x+3)} ?

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

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integral of 1/(sqrt(x+4)+\sqrt{x+3)}

\int \frac{1}{x + 4 + x + 3} d x

\frac{2}{3} (x + 4)^{\frac{3}{2}} - \frac{2}{3} (x + 3)^{\frac{3}{2}} + C

Solution steps

\int \frac{1}{x + 4 + x + 3} d x

\frac{1}{x + 4 + x + 3} = \frac{x + 4 - x + 3}{( x + 4 ) ^{2} - ( x + 3 ) ^{2}}

= \int \frac{x + 4 - x + 3}{( x + 4 ) ^{2} - ( x + 3 ) ^{2}} d x

Expand

\frac{x + 4 - x + 3}{( x + 4 ) ^{2} - ( x + 3 ) ^{2}} : \frac{x + 4}{( x + 4 ) ^{2} - ( x + 3 ) ^{2}} - \frac{x + 3}{( x + 4 ) ^{2} - ( x + 3 ) ^{2}}

Apply the Sum Rule:

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

= \int \frac{x + 4}{( x + 4 ) ^{2} - ( x + 3 ) ^{2}} d x - \int \frac{x + 3}{( x + 4 ) ^{2} - ( x + 3 ) ^{2}} d x

\int \frac{x + 4}{( x + 4 ) ^{2} - ( x + 3 ) ^{2}} d x = \frac{2}{3} (x + 4)^{\frac{3}{2}}

\int \frac{x + 3}{( x + 4 ) ^{2} - ( x + 3 ) ^{2}} d x = \frac{2}{3} (x + 3)^{\frac{3}{2}}

= \frac{2}{3} (x + 4)^{\frac{3}{2}} - \frac{2}{3} (x + 3)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{3} (x + 4)^{\frac{3}{2}} - \frac{2}{3} (x + 3)^{\frac{3}{2}} + C

y^{^{''}} + 4 y^{^{'}} + 8 y = 4 cos (2 t) + 2 sin (2 t)

\frac{d}{d x} (\frac{x ^{8}}{8 ( 1 - x ^{2} ) ^{4}})

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

\frac{\partial}{\partial y} (4 y^{3} - 4 x)

23 cos (x^{2})

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