Question
$\int\frac{\text{x}-1}{\sqrt{\text{x}+4}}\ \text{dx}$
✓
Answer
$\text{Let I} =\int\Big(\frac{\text{x}-1}{\sqrt{\text{x}+4}}\Big)\text{dx}$
Putting x + 4 = t
Then, x = t - 4
Difference both sides
dx = dt
Now integral becomes,
$\text{I}=\int\Big(\frac{\text{t}-4-1}{\sqrt{\text{t}}}\Big)\text{dt}$
$=\int\Big(\frac{\text{t}}{\sqrt{\text{t}}}-\frac{5}{\sqrt{\text{t}}}\Big)\text{dt}$
$=\int\Big(\text{t}^{\frac{1}{2}}-5\text{t}^{-\frac{1}{2}}\Big)\text{dt}$
$=\frac{\text{t}^{\frac{1}{2}+1}}{\frac{1}{2}+1}-5\frac{\text{t}^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+\text{C}$
$=\frac{2}{3}\text{t}^\frac{3}{2}-10\sqrt{\text{t}}+\text{C}$
$=\frac{2}{3}(\text{x}+4)^\frac{3}{2}-10(\text{x}+4)^\frac{1}{2}+\text{C}$
Putting x + 4 = t
Then, x = t - 4
Difference both sides
dx = dt
Now integral becomes,
$\text{I}=\int\Big(\frac{\text{t}-4-1}{\sqrt{\text{t}}}\Big)\text{dt}$
$=\int\Big(\frac{\text{t}}{\sqrt{\text{t}}}-\frac{5}{\sqrt{\text{t}}}\Big)\text{dt}$
$=\int\Big(\text{t}^{\frac{1}{2}}-5\text{t}^{-\frac{1}{2}}\Big)\text{dt}$
$=\frac{\text{t}^{\frac{1}{2}+1}}{\frac{1}{2}+1}-5\frac{\text{t}^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+\text{C}$
$=\frac{2}{3}\text{t}^\frac{3}{2}-10\sqrt{\text{t}}+\text{C}$
$=\frac{2}{3}(\text{x}+4)^\frac{3}{2}-10(\text{x}+4)^\frac{1}{2}+\text{C}$
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