Find out the answer to "Evaluate the following integrals. ∫13dtt(t+1)"

sibuzwaW

Answered question

2021-10-23

Evaluate the following integrals.

\int_{1}^{3} \frac{d t}{\sqrt{t} (t + 1)}

stuth1

Skilled2021-10-24Added 97 answers

Given :

\int_{1}^{3} \frac{1}{\sqrt{t} (t + 1)} d t

\int_{1}^{3} \frac{1}{\sqrt{t} (t + 1)} d t

put

u = \sqrt{t}

d u = \frac{1}{2 \sqrt{t}}

t = 1 \Rightarrow u = 1

t = 3 \Rightarrow u = \sqrt{3}

= \int_{1}^{3} \frac{1}{(t + 1) \sqrt{t}} d t

= 2 \cdot \int_{1}^{\sqrt{3}} \frac{1}{u^{2} + 1} d u

\int \frac{1}{u^{2} + 1} d u = \tan^{- 1} (u)

= 2 [\tan^{- 1} (\sqrt{3}) - \tan^{- 1} (1)]

= 2 [\frac{π}{3} - \frac{π}{4}]

= 2 \cdot \frac{π}{12}

= \frac{π}{6}

\int_{1}^{3} \frac{1}{\sqrt{t} (t + 1)} d t = \frac{π}{6}

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