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Integral with ax+b\sqrt{ax+b}

10​

∫ax+b dx=23a(ax+b)3+C\int \sqrt{ax+b}\ \mathrm{d}x=\frac{2}{3a}\sqrt{(ax+b)^3}+C
Derivation

Let u=ax+bu=ax+b, then du=a dx  ⟹  dx=dua\mathrm{d}u=a\ \mathrm{d}x \implies \mathrm{d}x=\frac{\mathrm{d}u}{a}

I=∫ax+b dx=∫ua du=u12+1a(12+1)+C=2u33a+C=23a(ax+b)3+C\begin{align*} I&=\int{\sqrt{ax+b}}\ \mathrm{d}x\\ &=\int{\frac{\sqrt{u}}{a}}\ \mathrm{d}u\\ &=\frac{u^{\frac{1}{2}+1}}{a(\frac{1}{2}+1)}+C\\ &=\frac{2\sqrt{u^3}}{3a}+C\\ &=\frac{2}{3a}\sqrt{(ax+b)^3}+C \tag*{$\blacksquare$} \end{align*}

11​

∫xax+b  dx=215a2(3ax−2b)(ax+b)3+C\int x\sqrt {ax+b}\ \ \mathrm{d}x=\frac{2}{15a^2}(3ax-2b)\sqrt{(ax+b)^3}+C
Derivation

Let u=ax+bu=\sqrt{ax+b}, then x=u2−bax=\frac{u^2-b}{a}, du=a2ax+bdx=a2udx  ⟹  dx=2uadu\mathrm{d}u=\frac{a}{2\sqrt{ax+b}}\mathrm{d}x=\frac{a}{2u}{\mathrm{d}}x \implies \mathrm{d}x=\frac{2u}{a}{\mathrm{d}}u

I=∫xax+b dx=∫u2−ba⋅u⋅2ua du(Back Substitution)=2a2∫(u4−bu2) du=2a2(15u5−b3u3+C)=2a2(3u5−5bu315+C)=2a2[(3u2−5b)u315+C]=215a2[3(ax+b)2−5b](ax+b)3+C(plug in u=ax+b)=215a2(3ax−2b)(ax+b)3+C\begin{align*} I&=\int{x\sqrt{ax+b}}\ \mathrm{d}x\\ &=\int\frac{u^2-b}{a}\cdot u\cdot \frac{2u}{a}\ \mathrm{d}u &&\text{(Back Substitution)}\\ &=\frac{2}{a^2}\int{(u^4-bu^2)}\ \mathrm{d}u\\ &=\frac{2}{a^2}\left(\frac{1}{5}u^5-\frac{b}{3}u^3 +C\right)\\ &=\frac{2}{a^2}\left(\frac{3u^5-5bu^3}{15}+C\right)\\ &=\frac{2}{a^2}\left[\frac{(3u^2-5b)u^3}{15}+C\right]\\ &=\frac{2}{15a^2}\left[3(\sqrt{ax+b})^2-5b\right]\sqrt{(ax+b)^3}+C &&(\text{plug in }u=\sqrt{ax+b})\\ &=\frac{2}{15a^2}(3ax-2b)\sqrt{(ax+b)^3}+C \tag*{$\blacksquare$} \end{align*}

Back Substitution

12​

∫x2ax+b  dx=2105a3(15a2x2−12abx+8b2)(ax+b)3+C\int x^2\sqrt {ax+b}\ \ \mathrm{d}x= \frac{2}{105a^3}(15a^2x^2-12abx+8b^2)\sqrt{(ax+b)^3}+C
Derivation

Let u=ax+bu=\sqrt{ax+b}, then x=u2−bax=\frac{u^2-b}{a}, du=a2ax+bdx=a2udx  ⟹  dx=2uadu\mathrm{d}u=\frac{a}{2\sqrt{ax+b}}\mathrm{d}x=\frac{a}{2u}{\mathrm{d}}x \implies \mathrm{d}x=\frac{2u}{a}{\mathrm{d}}u

I=∫x2ax+b dx=∫(u2−ba)2⋅u⋅2ua du(Back Substitution)=2a3∫(u4−2bu2+b2)(u2) du=2a3∫(u6−2bu4+b2u2) du=2a2(17u7−2b5u5+b23u3+C)=2a2(15u7−42bu5+35b2u3105+C)=2a2[(15u4−42bu2+35b2)u3105+C]=2105a3[15(ax+b)4−42(ax+b)2+35b2](ax+b)3+C(plug in u=ax+b)=2105a3(15a2x2−12abx+8b2)(ax+b)3+C\begin{align*} I&=\int{x^2\sqrt{ax+b}}\ \mathrm{d}x\\ &=\int{\left(\frac{u^2-b}{a}\right)^2\cdot u\cdot \frac{2u}{a}}\ \mathrm{d}u &&\text{(Back Substitution)}\\ &=\frac{2}{a^3}\int{(u^4-2bu^2+b^2)(u^2)}\ \mathrm{d}u\\ &=\frac{2}{a^3}\int{(u^6-2bu^4+b^2u^2)}\ \mathrm{d}u\\ &=\frac{2}{a^2}\left(\frac{1}{7}u^7-\frac{2b}{5}u^5+\frac{b^2}{3}u^3 +C\right)\\ &=\frac{2}{a^2}\left(\frac{15u^7-42bu^5+35b^2u^3}{105}+C\right)\\ &=\frac{2}{a^2}\left[\frac{(15u^4-42bu^2+35b^2)u^3}{105}+C\right]\\ &=\frac{2}{105a^3}\left[15(\sqrt{ax+b})^4-42(\sqrt{ax+b})^2+35b^2\right]\sqrt{(ax+b)^3}+C &&(\text{plug in }u=\sqrt{ax+b})\\ &=\frac{2}{105a^3}(15a^2x^2-12abx+8b^2)\sqrt{(ax+b)^3}+C \tag*{$\blacksquare$} \end{align*}

Back Substitution

13​

∫xax+b dx=23a2(ax−2b)ax+b+C\int \frac{x}{\sqrt{ax+b}}\ \mathrm{d}x=\frac{2}{3a^2}(ax-2b)\sqrt{ax+b}+C
Derivation

Let u=ax+bu=\sqrt{ax+b}, then x=u2−bax=\frac{u^2-b}{a}, du=a2ax+bdx=a2udx  ⟹  dx=2uadu\mathrm{d}u=\frac{a}{2\sqrt{ax+b}}\mathrm{d}x=\frac{a}{2u}{\mathrm{d}}x \implies \mathrm{d}x=\frac{2u}{a}{\mathrm{d}}u

I=∫xax+b dx=∫u2−bau2ua du(Back Substitution)=∫2(u2−b)a2 du=2a2∫(u2−b) du=2a2(13u3−bu+C)=23a2⋅(u2−3b)⋅u+C=23a2⋅(ax+b−3b)⋅ax+b+C(plug in u=ax+b)=23a2(ax−2b)ax+b+C\begin{align*} I&=\int{\frac{x}{\sqrt{ax+b}}}\ \mathrm{d}x\\ &=\int{\frac{\frac{u^2-b}{a}}{u}}\frac{2u}{a}\ \mathrm{d}u &&\text{(Back Substitution)}\\ &=\int{\frac{2(u^2-b)}{a^2}}\ \mathrm{d}u\\ &=\frac{2}{a^2}\int{(u^2-b)}\ \mathrm{d}u\\ &=\frac{2}{a^2}\left(\frac{1}{3}u^3-bu+C\right)\\ &=\frac{2}{3a^2}\cdot (u^2-3b)\cdot u+C\\ &=\frac{2}{3a^2}\cdot(ax+b-3b)\cdot \sqrt{ax+b}+C &&(\text{plug in }u=\sqrt{ax+b})\\ &=\frac{2}{3a^2}(ax-2b)\sqrt{ax+b}+C \tag*{$\blacksquare$} \end{align*}

Back Substitution

14​

∫x2ax+b dx=215a3(3a2x2−4abx+8b2)ax+b+C\int\frac{x^2}{\sqrt{ax+b}}\ \mathrm{d}x=\frac{2}{15a^3}\left(3a^2x^2-4abx+8b^2\right)\sqrt{ax+b}+C
Derivation

Let u=ax+bu=\sqrt{ax+b}, then x=u2−bax=\frac{u^2-b}{a}, du=a2ax+bdx=a2udx  ⟹  dx=2uadu\mathrm{d}u=\frac{a}{2\sqrt{ax+b}}\mathrm{d}x=\frac{a}{2u}{\mathrm{d}}x \implies \mathrm{d}x=\frac{2u}{a}{\mathrm{d}}u

I=∫x2ax+b dx=∫(u2−ba)2u2ua du(Back Substitution)=2a3∫(u2−b)2 du=2a3∫(u4−2bu2+b2) du=2a3(15u5−2b3u3+b2u+C)=2a3⋅(3u4−10bu2+15b215)⋅u+C=215a3⋅[3(ax+b)2−10b(ax+b)+15b2]⋅ax+b+C(plug in u=ax+b)=215a3(3a2x2−4abx+8b2)ax+b+C\begin{align*} I&=\int\frac{x^2}{\sqrt{ax+b}}\ \mathrm{d}x\\ &=\int\frac{\left(\frac{u^2-b}{a} \right)^2}{u}\frac{2u}{a}\ \mathrm{d}u &&\text{(Back Substitution)}\\ &=\frac{2}{a^3}\int{(u^2-b)^2}\ \mathrm{d}u\\ &=\frac{2}{a^3}\int(u^4-2bu^2+b^2)\ \mathrm{d}u\\ &=\frac{2}{a^3}\left(\frac{1}{5}u^5-\frac{2b}{3}u^3+b^2u+C\right)\\ &=\frac{2}{a^3}\cdot\left(\frac{3u^4-10bu^2+15b^2}{15}\right)\cdot u+C\\ &=\frac{2}{15a^3}\cdot \left[3(ax+b)^2-10b(ax+b)+15b^2\right]\cdot \sqrt{ax+b}+C &&(\text{plug in }u=\sqrt{ax+b})\\ &=\frac{2}{15a^3}(3a^2x^2-4abx+8b^2)\sqrt{ax+b}+C \tag*{$\blacksquare$} \end{align*}

Back Substitution

15​

∫1xax+b dx={1bln⁡∣ax+b−bax+b+b∣+C(b>0)2−barctan⁡ax+b−b+C(b<0)\int \frac{1}{x \sqrt{ax+b}} \ \mathrm{d}x = \left\{ \begin{aligned} &\frac{1}{\sqrt{b}} \ln{\left|\frac{\sqrt{ax+b}-\sqrt{b}}{\sqrt{ax+b}+\sqrt{b}}\right|} + C \quad (b>0)\\ &\frac{2}{\sqrt{-b}} \arctan{\sqrt{\frac{ax+b}{-b}}}+C \quad (b<0) \end{aligned} \right.
Derivation

Let u=ax+bu=\sqrt{ax+b}, then x=u2−bax=\frac{u^2-b}{a}, du=a2ax+bdx=a2udx  ⟹  dx=2uadu\mathrm{d}u=\frac{a}{2\sqrt{ax+b}}\mathrm{d}x=\frac{a}{2u}{\mathrm{d}}x \implies \mathrm{d}x=\frac{2u}{a}{\mathrm{d}}u

I=∫1xax+b dx=∫1u2−ba⋅u⋅2ua du(Back Substitution)=2∫1u2−b du\begin{align*} I&=\int{\frac{1}{x\sqrt{ax+b}}}\ \mathrm{d}x\\ &=\int{\frac{1}{\frac{u^2-b}{a}\cdot u}}\cdot \frac{2u}{a}\ \mathrm{d}u &&\text{(Back Substitution)} \\ &=2\int{\frac{1}{u^2-b}}\ \mathrm{d}u \end{align*}

When b>0b>0 :

2∫1u2−b du=2∫1u2−(b)2 du=1bln⁡∣u−bu+b∣+C(Integral Formula #21)=1bln⁡∣ax+b−bax+b+b∣+C(plug in u=ax+b)\begin{align*} 2\int{\frac{1}{u^2-b}}\ \mathrm{d}u&=2\int{\frac{1}{u^2-(\sqrt{b})^2}}\ \mathrm{d}u\\ &=\frac{1}{\sqrt{b}}\ln{\left|\frac{u-\sqrt{b}}{u+\sqrt{b}}\right|}+C &&(\text{Integral Formula \#21})\\ &=\frac{1}{\sqrt{b}}\ln{\left|\frac{\sqrt{ax+b}-\sqrt{b}}{\sqrt{ax+b}+\sqrt{b}}\right|}+C &&(\text{plug in }u=\sqrt{ax+b}) \end{align*}

When b<0b<0 :

2∫1u2−b du=2∫1u2+(−b)2 du=2[1−barctan⁡(u−b)+C](Integral Formula #19)=2−barctan⁡ax+b−b+C(plug in u=ax+b)\begin{align*} 2\int{\frac{1}{u^2-b}}\ \mathrm{d}u&=2\int{\frac{1}{u^2+(\sqrt{-b})^2}}\ \mathrm{d}u\\ &=2\left[\frac{1}{\sqrt{-b}}\arctan{\left(\frac{u}{\sqrt{-b}}\right)}+C\right] &&(\text{Integral Formula \#19})\\ &=\frac{2}{\sqrt{-b}}\arctan{\sqrt{\frac{ax+b}{-b}}}+C &&(\text{plug in }u=\sqrt{ax+b}) \end{align*}

Hence:

∫1xax+b dx={1bln⁡∣ax+b−bax+b+b∣+C (b>0)2−barctan⁡ax+b−b+C (b<0)■\int \frac{1}{x \sqrt{ax+b}} \ \mathrm{d}x = \left\{ \begin{aligned} &\frac{1}{\sqrt{b}} \ln{\left|\frac{\sqrt{ax+b}-\sqrt{b}}{\sqrt{ax+b}+\sqrt{b}}\right|} + C \ \left(b>0\right)\\ &\frac{2}{\sqrt{-b}} \arctan{\sqrt{\frac{ax+b}{-b}}}+C \ \left(b<0\right) \end{aligned} \right. \tag*{$\blacksquare$}

Back Substitution

Integral Formula #21

Integral Formula #19

16​

∫ax+bx2 dx=−ax+bx+a2∫1xax+b dx\int{\frac{\sqrt{ax+b}}{x^2}}\ \mathrm{d}x=-\frac{\sqrt{ax+b}}{x}+\frac{a}{2}\int{\frac{1}{x\sqrt{ax+b}}}\ \mathrm{d}x
Derivation
I=∫ax+bx2 dx=∫ax+b⋅1x2 dx=∫ax+b d(−1x)=−ax+bx−∫−1x d(ax+b)(Integration by Parts)=−ax+bx+a2∫1xax+b dx\begin{align*} I&=\int{\frac{\sqrt{ax+b}}{x^2}}\ \mathrm{d}x\\ &=\int{\sqrt{ax+b}}\cdot \frac{1}{x^2}\ \mathrm{d}x\\ &=\int{\sqrt{ax+b}}\ \mathrm{d}\left(-\frac{1}{x}\right)\\ &=-\frac{\sqrt{ax+b}}{x}-\int{-\frac{1}{x}\ \mathrm{d}(\sqrt{ax+b})} &&(\text{Integration by Parts})\\ &=-\frac{\sqrt{ax+b}}{x}+\frac{a}{2}\int{\frac{1}{x\sqrt{ax+b}}}\ \mathrm{d}x \tag*{$\blacksquare$} \end{align*}

[Integration by Parts](link tbd)

17​

∫ax+bx dx=2ax+b+b∫1xax+b dx\int\frac{\sqrt{ax+b}}{x}\ \mathrm{d}x=2\sqrt{ax+b}+b\int{\frac{1}{x\sqrt{ax+b}}}\ \mathrm{d}x
Derivation

Let u=ax+bu=\sqrt{ax+b}, then x=u2−bax=\frac{u^2-b}{a}, du=a2ax+bdx=a2udx  ⟹  dx=2uadu\mathrm{d}u=\frac{a}{2\sqrt{ax+b}}\mathrm{d}x=\frac{a}{2u}{\mathrm{d}}x \implies \mathrm{d}x=\frac{2u}{a}{\mathrm{d}}u

I=∫ax+bx dx=∫uu2−ba⋅2ua du(Back Substitution)=∫2u2u2−b du=2∫u2−b+bu2−b du=2∫(1+bu2−b) du=2u+2∫bu2−b du=2ax+b+2∫bax+b−b d(ax+b)(plug in u=ax+b)=2ax+b+2∫bax⋅a2ax+b dx=2ax+b+b∫1xax+b dx\begin{align*} I&=\int{\frac{\sqrt{ax+b}}{x}}\ \mathrm{d}x\\ &=\int{\frac{u}{\frac{u^2-b}{a}}\cdot \frac{2u}{a}}\ \mathrm{d}u &&(\text{Back Substitution})\\ &=\int{\frac{2u^2}{u^2-b}}\ \mathrm{d}u\\ &=2\int{\frac{u^2-b+b}{u^2-b}}\ \mathrm{d}u\\ &=2\int{\left(1+\frac{b}{u^2-b} \right)}\ \mathrm{d}u\\ &=2u+2\int{\frac{b}{u^2-b}}\ \mathrm{d}u\\ &=2\sqrt{ax+b}+2\int{\frac{b}{ax+b-b}}\ \mathrm{d}(\sqrt{ax+b}) &&(\text{plug in }u=\sqrt{ax+b})\\ &=2\sqrt{ax+b}+2\int{\frac{b}{ax}}\cdot \frac{a}{2\sqrt{ax+b}}\ \mathrm{d}x\\ &=2\sqrt{ax+b}+b\int{\frac{1}{x\sqrt{ax+b}}}\ \mathrm{d}x \tag*{$\blacksquare$} \end{align*}

Back Substitution

18​

∫1x2ax+b dx=−ax+bbx−a2b∫1xax+b dx\int\frac{1}{x^2\sqrt{ax+b}}\ \mathrm{d}x=-\frac{\sqrt{ax+b}}{bx}-\frac{a}{2b}\int\frac{1}{x\sqrt{ax+b}}\ \mathrm{d}x
Derivation
I=∫1x2ax+b dx=∫(Axax+b+Bax+bx2) dx(Partial Fraction Decomposition)=∫Ax+B(ax+b)x2ax+b dx=∫(A+Ba)x+Bbx2ax+b dx\begin{align*} I&=\int\frac{1}{x^2\sqrt{ax+b}}\ \mathrm{d}x\\ &=\int{\left(\frac{A}{x\sqrt{ax+b}}+\frac{B\sqrt{ax+b}}{x^2}\right)}\ \mathrm{d}x &&\text{(Partial Fraction Decomposition)}\\ &=\int{\frac{Ax+B(ax+b)}{x^2\sqrt{ax+b}}}\ \mathrm{d}x\\ &=\int{\frac{(A+Ba)x+Bb}{x^2\sqrt{ax+b}}}\ \mathrm{d}x \end{align*}

Comparing coefficient gives:

(A+Ba)x+Bb≡1  ⟹  {A+Ba=0Bb=1  ⟹  {A=−abB=1b(A+Ba)x+Bb\equiv 1 \implies \left\{ \begin{align*} &A+Ba=0\\ &Bb=1 \end{align*} \right. \implies \left\{ \begin{align*} &A=-\frac{a}{b}\\ &B=\frac{1}{b} \end{align*} \right.

Plug in back:

I=∫(−abxax+b+1bax+bx2) dx=−ab∫1xax+b dx+1b∫ax+bx2 dx=−ab∫1xax+b dx+1b(−ax+bx+a2∫1xax+b dx)(Integral Formula #16)=−ax+bbx+(−ab+a2b)∫1xax+b dx=−ax+bbx−a2b∫1xax+b dx\begin{align*} I&=\int{\left(\frac{-\frac{a}{b}}{x\sqrt{ax+b}}+\frac{\frac{1}{b}\sqrt{ax+b}}{x^2}\right)}\ \mathrm{d}x\\ &=-\frac{a}{b}\int{\frac{1}{x\sqrt{ax+b}}}\ \mathrm{d}x+\frac{1}{b}\int{\frac{\sqrt{ax+b}}{x^2}}\ \mathrm{d}x\\ &=-\frac{a}{b}\int{\frac{1}{x\sqrt{ax+b}}}\ \mathrm{d}x+\frac{1}{b}\left(-\frac{\sqrt{ax+b}}{x}+\frac{a}{2}\int{\frac{1}{x\sqrt{ax+b}}}\ \mathrm{d}x\right) &&(\text{Integral Formula \#16})\\ &=-\frac{\sqrt{ax+b}}{bx}+\left(-\frac{a}{b}+\frac{a}{2b}\right)\int{\frac{1}{x\sqrt{ax+b}}}\ \mathrm{d}x\\ &=-\frac{\sqrt{ax+b}}{bx}-\frac{a}{2b}\int\frac{1}{x\sqrt{ax+b}}\ \mathrm{d}x \tag*{$\blacksquare$} \end{align*}

Integral Formula #16