How do you integrate by substitution $\int \frac{t + 2 t^{2}}{\sqrt{t}} d t$ $int (t+2t^2)/sqrtt dt$ ?

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How do you integrate by substitution $\int \frac{t + 2 t^{2}}{\sqrt{t}} d t$ $int (t+2t^2)/sqrtt dt$ ?

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Answer 1 · 2017-12-26T23:06:10

$\int \frac{t + 2 t^{2}}{\sqrt{t}} d t = 2 \frac{\sqrt{t^{3}}}{3} + 4 \frac{\sqrt{t^{5}}}{5} + C$ $int(t+2t^2)/sqrt(t)dt=2sqrt(t^3)/3+4sqrt(t^5)/5+C$

Explanation:

We see a $\sqrt{t}$ $sqrt(t)$ our first thought is to substitude $u = \sqrt{t}$ $u=sqrt(t)$ .

So let's do that :

$u = \sqrt{t} \Rightarrow d u = \frac{1}{2 \sqrt{t}} d t \Rightarrow d t = 2 \sqrt{t} d u \Rightarrow d t = 2 u d u$ $u=sqrt(t)=>du=1/(2sqrt(t))dt=>dt=2sqrt(t)du=>dt=2udu$

So our integral becomes :

$\int \frac{t + 2 t^{2}}{\sqrt{t}} d t = \int \frac{u^{2} + 2 u^{4}}{u} \cdot 2 u d u = 2 \int (u^{2} + 2 u^{4}) d u =$ $int(t+2t^2)/sqrt(t)dt=int(u^2+2u^4)/u*2udu=2int(u^2+2u^4)du=$

$2 (\frac{u^{3}}{3} + 2 \frac{u^{5}}{5}) + C = 2 \frac{u^{3}}{3} + 4 \frac{u^{5}}{5} + C$ $2(u^3/3+2u^5/5)+C=2u^3/3+4u^5/5+C$

Now let's substitude $u = \sqrt{t}$ $u=sqrt(t)$ back in :

$\int \frac{t + 2 t^{2}}{\sqrt{t}} d t = 2 \frac{\sqrt{t^{3}}}{3} + 4 \frac{\sqrt{t^{5}}}{5} + C$ $int(t+2t^2)/sqrt(t)dt=2sqrt(t^3)/3+4sqrt(t^5)/5+C$

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How do you integrate by substitution $\int \frac{t + 2 t^{2}}{\sqrt{t}} d t$ $int (t+2t^2)/sqrtt dt$ ?

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