Multi-method for solving definite integral

Integration by parts

∫f(x)g′(x)dx=f(x)g(x)−∫f′(x)g(x)⋅dx⇓∫abf(x)g′(x)dx=f(x)g(x)∣ab−∫abf′(x)g(x)⋅dx\begin{align*} \\ \\ \int f(x)g^\prime(x)dx &= f(x)g(x) - \int f^\prime(x)g(x) \cdot dx \\ \\ &\Downarrow \\ \\ \int_a^b f(x)g^\prime(x)dx &= f(x)g(x)|_a^b - \int_a^b f^\prime(x)g(x) \cdot dx \end{align*}

Substitution method

∫1215x−1⋅dxt=5x−1t2=5x−1t2+15=xd[15(t2+1)]=dx15(2t)dt=dxx∣1→2t∣5⋅1−1→5⋅2−1t∣2→3∫231t⋅(15(2t)dt)=∫2325⋅dt=(25⋅t)∣23=65−45=25\begin{align*} &\int_1^2 \frac{1}{\sqrt{5x-1}} \cdot dx \\ \\ &t = \sqrt{5x-1} \\ \\ & t^2 = 5x-1 \\ \\ &\frac{t^2 + 1}{5} = x \\ \\ &d[\frac{1}{5} (t^2 + 1)] = dx \\ \\ &\frac{1}{5} (2t) dt = dx \\ \\ & x | 1 \rightarrow 2 \\ & t | \sqrt{5 \cdot 1 -1} \rightarrow \sqrt{5 \cdot 2 -1} \\ & t | 2 \rightarrow 3 \\ \\ &\int_2^3 \frac{1}{t} \cdot (\frac{1}{5} (2t) dt) \\ \\ &= \int_2^3 \frac{2}{5} \cdot dt \\ \\ &= (\frac{2}{5} \cdot t)|_2^3 \\ \\ &= \frac{6}{5} - \frac{4}{5} \\ \\ &= \frac{2}{5} \end{align*}
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