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

Evaluate the following integral.
$$\displaystyle\int{t}\sqrt{{{t}^{{{2}}}+{1}}}{\left.{d}{t}\right.}$$

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Tasneem Almond
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 $$\displaystyle{u}={t}^{{{2}}}+{1}$$,
$$\displaystyle={\frac{{{1}}}{{{3}}}}{\left({t}^{{{2}}}+{1}\right)}^{{{\frac{{{3}}}{{{2}}}}}}+{C}$$