Evaluate the following integrals. int(x/(sqrt(x-4))dx

Question
Applications of integrals
Evaluate the following integrals.
$$\displaystyle\int{\left(\frac{{x}}{{\sqrt{{{x}-{4}}}}}{\left.{d}{x}\right.}\right.}$$

2021-01-14
Step 1
Consider the integrals,
$$\displaystyle\int{\left(\frac{{x}}{{\sqrt{{{x}-{4}}}}}{\left.{d}{x}\right.}\right.}$$
Suppose that, $$\displaystyle\sqrt{{{x}-{4}}}={t}$$
Differentiating with respect to "x"
$$\displaystyle\frac{{1}}{{{2}\sqrt{{{x}-{4}}}}}{\left.{d}{x}\right.}={\left.{d}{t}\right.}$$
$$\displaystyle\frac{{1}}{{\sqrt{{{x}-{4}}}}}{\left.{d}{x}\right.}={2}{\left.{d}{t}\right.}$$
Step 2
Substitute all value in given integrals,
$$\displaystyle\int\frac{{x}}{{\sqrt{{{x}-{4}}}}}{\left.{d}{x}\right.}=\int{\left({t}^{{2}}+{4}\right)}{2}{\left.{d}{t}\right.}$$
$$\displaystyle={2}\int{\left({t}^{{2}}+{4}\right)}{\left.{d}{t}\right.}$$
$$\displaystyle={2}{\left[\frac{{t}^{{3}}}{{3}}+{4}{t}\right]}+{C}$$
$$\displaystyle=\frac{{{2}{t}^{{3}}}}{{3}}+{8}{t}+{C}$$
$$\displaystyle=\frac{{2}}{{3}}{\left({x}-{4}\right)}^{{\frac{{3}}{{2}}}}+{8}\sqrt{{{x}-{4}}}+{C}$$

Relevant Questions

Evaluate the following integrals.
$$\displaystyle\int{\frac{{{\left.{d}{x}\right.}}}{{\sqrt{{{\left({x}-{1}\right)}{\left({3}-{x}\right)}}}}}}$$
Evaluate the following integrals.
$$\displaystyle\int\frac{{{\left.{d}{x}\right.}}}{{\sqrt{{{\left({x}-{1}\right)}{\left({3}-{x}\right)}}}}}$$
Evaluate the following integral.
$$\displaystyle\int{\frac{{{x}}}{{\sqrt{{{x}-{4}}}}}}{\left.{d}{x}\right.}$$
Evaluate the integrals.
$$\displaystyle\int{\frac{{{{\sin}^{{-{1}}}{x}}}}{{\sqrt{{{1}-{x}^{{{2}}}}}}}}{\left.{d}{x}\right.}$$
Evaluate the following integrals.
$$\displaystyle\int{\left({2}{x}^{{{3}}}-{x}^{{{2}}}+{3}{x}-{7}\right)}{\left.{d}{x}\right.}$$
Evaluate the following integrals.
$$\displaystyle\int\frac{{{\left.{d}{x}\right.}}}{{{x}^{{3}}-{x}^{{2}}}}$$
Evaluate the following integrals.
$$\displaystyle{\int_{{-{2}}}^{{-{1}}}}\sqrt{{-{4}{x}-{x}^{{{2}}}}}{\left.{d}{x}\right.}$$
Convert the above indefinite integrals into definate integrals using the intervals $$\displaystyle{\left[{0},{1}\right]}$$.
(a)$$\displaystyle\int\sqrt{{{a}^{{2}}-{x}^{{2}}}}{\left.{d}{x}\right.}$$
(b)$$\displaystyle\int\sqrt{{{1}-{x}^{{2}}}}{\left.{d}{x}\right.}$$
Which of the following integrals are improper integrals?
1.$$\displaystyle{\int_{{{0}}}^{{{3}}}}{\left({3}-{x}\right)}^{{{2}}}{\left\lbrace{3}\right\rbrace}{\left.{d}{x}\right.}$$
2.$$\displaystyle{\int_{{{1}}}^{{{16}}}}{\frac{{{e}^{{\sqrt{{{x}}}}}}}{{\sqrt{{{x}}}}}}{\left.{d}{x}\right.}$$
3.$$\displaystyle{\int_{{{1}}}^{{\propto}}}{\frac{{{3}}}{{\sqrt{{{3}}}{\left\lbrace{x}\right\rbrace}}}}{\left.{d}{x}\right.}$$
4.$$\displaystyle{\int_{{-{2}}}^{{{2}}}}{3}{\left({x}+{1}\right)}^{{-{1}}}{\left.{d}{x}\right.}$$
a) 1 only
b)1 and 2
c)3 only
d)2 and 3
e)1,3 and 4
f)All of the integrals are improper
$$\displaystyle\int{x}{{\sin}^{{-{{1}}}}{2}}{x}{\left.{d}{x}\right.}$$