Evaluate the following integral. \int t\sqrt{t^{2}+1}dt

facas9 2021-09-22 Answered
Evaluate the following integral.
\(\displaystyle\int{t}\sqrt{{{t}^{{{2}}}+{1}}}{\left.{d}{t}\right.}\)

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Expert Answer

Tasneem Almond
Answered 2021-09-23 Author has 11293 answers
Step 1
Consider the integrals:
\(\displaystyle\int{t}\sqrt{{{t}^{{{2}}}+{1}}}{\left.{d}{t}\right.}\)...(1)
Let,
\(\displaystyle{u}={t}^{{{2}}}+{1}\)
Differentiating   w.r.t.t
du=2tdt+0
du=2tdt
\(\displaystyle{t}\cdot{\left.{d}{t}\right.}={\frac{{{1}}}{{{2}}}}{d}{u}\)
Step 2
Substitute \(\displaystyle{u}={t}^{{{2}}}+{1}\) and \(\displaystyle{t}\cdot{\left.{d}{t}\right.}={\frac{{{1}}}{{{2}}}}{d}{u}\) in equation (1)
then,
\(\displaystyle\int{t}\sqrt{{{t}^{{{2}}}+{1}}}{\left.{d}{t}\right.}=\int\sqrt{{{u}}}\cdot{\frac{{{1}}}{{{2}}}}{d}{u}\)
\(\displaystyle={\frac{{{1}}}{{{2}}}}\int{u}^{{{\frac{{{1}}}{{{2}}}}}}{d}{u}\)
\(\displaystyle={\frac{{{1}}}{{{2}}}}{\left[{\frac{{{u}^{{{\frac{{{1}}}{{{2}}}}+{1}}}}}{{{\frac{{{1}}}{{{2}}}}+{1}}}}\right]}+{C}\)
\(\displaystyle={\frac{{{1}}}{{{2}}}}{\left[{\frac{{{2}{u}^{{{\frac{{{3}}}{{{2}}}}}}}}{{{3}}}}\right]}+{C}\)
\(\displaystyle={\frac{{{1}}}{{{3}}}}{u}^{{{\frac{{{3}}}{{{2}}}}}}+{C}\)
Back  substitute \(\displaystyle{u}={t}^{{{2}}}+{1}\),
\(\displaystyle={\frac{{{1}}}{{{3}}}}{\left({t}^{{{2}}}+{1}\right)}^{{{\frac{{{3}}}{{{2}}}}}}+{C}\)
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