solve this C: Calculus:Integral: Substitution:
Answers
\int \frac{5+\sqrt{x}}{2-\sqrt{x}} \, dx∫
2−√
x
, x={u}^{2}x=u
2
, and dx=2u \, dudx=2udu
2 Substitute variables from above.
\int \frac{5+u}{2-u}\times 2u \, du∫
2−u
5+u
×2udu
3 Use Constant Factor Rule: \int cf(x) \, dx=c\int f(x) \, dx∫cf(x)dx=c∫f(x)dx.
2\int \frac{u(5+u)}{2-u} \, du2∫
2−u
u(5+u)
du
4 Expand \frac{u(5+u)}{2-u}
2−u
u(5+u)
.
2\int \frac{5u+{u}^{2}}{2-u} \, du2∫
2−u
5u+u
2
du
5 Polynomial Division
2\int -u-7+\frac{14}{2-u} \, du2∫−u−7+
2−u
14
du
6 Use Sum Rule: \int f(x)+g(x) \, dx=\int f(x) \, dx+\int g(x) \, dx∫f(x)+g(x)dx=∫f(x)dx+∫g(x)dx.
2(\int -u-7 \, du+\int \frac{14}{2-u} \, du)2(∫−u−7du+∫
2−u
14
du)
7 Use Power Rule: \int {x}^{n} \, dx=\frac{{x}^{n+1}}{n+1}+C∫x
n
dx=
n+1
x
n+1
+C.
2(-\frac{{u}^{2}}{2}-7u+\int \frac{14}{2-u} \, du)2(−
2
u
2
−7u+∫
2−u
14
du)
8 Use Constant Factor Rule: \int cf(x) \, dx=c\int f(x) \, dx∫cf(x)dx=c∫f(x)dx.
2(-\frac{{u}^{2}}{2}-7u+14\int \frac{1}{2-u} \, du)2(−
2
u
2
−7u+14∫
2−u
1
du)
9 Use Integration by Substitution on \int \frac{1}{2-u} \, du∫
2−u
1
du.
Let w=2-uw=2−u, dw=-dudw=−du
10 Using ww and dwdw above, rewrite \int \frac{1}{2-u} \, du∫
2−u
1
du.
\int -\frac{1}{w} \, dw∫−
w
1
dw
11 The derivative of \ln{x}lnx is \frac{1}{x}
x
1
.
-\ln{w}−lnw
12 Substitute w=2-uw=2−u back into the original integral.
-\ln{(2-u)}−ln(2−u)
13 Rewrite the integral with the completed substitution.
2(-\frac{{u}^{2}}{2}-7u-14\ln{(2-u)})2(−
2
u
2
−7u−14ln(2−u))
14 Expand.
-{u}^{2}-14u-28\ln{(2-u)}−u
2
−14u−28ln(2−u)
15 Substitute u=\sqrt{x}u=√
x
back into the original integral.
-{\sqrt{x}}^{2}-14\sqrt{x}-28\ln{(2-\sqrt{x})}−√
x
2
−14√
x
−28ln(2−√
x
)
16 Add constant.
-x-14\sqrt{x}-28\ln{(2-\sqrt{x})}+C−x−14√
x
−28ln(2−√
x
)+C
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